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CHEMISTRY, M7 2024 HSC 35

Unknown samples of three carboxylic acids, labelled \(\text{X , Y}\) and \(\text{Z}\), are analysed to determine their identities.

  • Both \(\text{Y}\) and \(\text{Z}\) react rapidly with bromine in the absence of UV light, but \(\text{X}\) does not. A 0.100 g sample of \(\text{Y}\) reacts with the same amount of bromine as a 0.200 g sample of \(\text{Z}\).
  • Separate 0.100 g samples of \(\text{X , Y}\) and \(\text{Z}\) are titrated with 0.0617 mol L\(^{-1}\) sodium hydroxide solution. The titre volumes are shown.

\(
\begin{array}{|l|c|c|c|}
\hline \textit{Acid } & X & Y & Z \\
\hline \begin{array}{l}
\text {Volume of } \ce{NaOH \text{(mL)}} \\
\end{array} & 21.88 & 22.49 & 22.49 \\
\hline
\end{array}
\)
 

  • Both \(\text{Y}\) and \(\text{Z}\) can undergo hydration reactions in the presence of a suitable catalyst. Two products are possible for the hydration of \(\text{Y}\), but only one product is possible with \(\text{Z}\).

Identify which structures 1, 2 and 3 in the table are acids \(\text{X , Y}\) and \(\text{Z}\). Justify your answer with reference to the information provided.   (7 marks)

   

Show Answers Only

→ Both sample \(\text{Y}\) and \(\text{Z}\) undergo an addition reaction with bromine and a hydration reaction. Therefore these samples must contain a \(\ce{C=C}\) bond.

→ As sample \(\text{X}\), undergoes neither of these reactions, it must have no \(\ce{C=C}\) bond, thus sample \(\text{X}\) is structure \(2\).

→ Both structure \(1\) and structure \(3\) contain 1 \(\ce{C=C}\) each \(\Rightarrow\) they will react in a \(1:1\) with \(\ce{Br2}\). Hence the same number of moles of the carboxylic acid samples will react with the bromine.

→ Since the mass of sample \(\text{Z}\) that reacts with the bromine is double the mass of sample \(\text{Y}\), the molar mass of sample \(\text{Z}\) must be double the molar mass of sample \(\text{Y}\) following the formula \(m = n \times MM\).

→ Therefore, sample \(\text{Y}\) is structure 1 and sample \(\text{Z}\) is structure 3.
 

Other information provided that could support identification includes:

→ The two products formed for the hydration of \(\text{Y}\) is due to the asymmetry of structure 1 and the single product formed in the hydration of \(\text{Z}\) is due to the symmetrical nature of structure 3.

The titration values are consistent with the proposed samples and their corresponding structures.

→ \(\ce{n(NaOH)}\) reacted with \(\text{X} = 0.0617 \times 0.02188 = 1.35 \times 10^{-3}\ \text{mol}\)

\(\text{n}_{\text{X}}= \dfrac{0.100}{74.078}= 0.00135\ \text{mol}\). Therefore \(\text{X}\) reacts in a \(1:1\) molar ratio as it is a monoprotic acid.

→ \(\ce{n(NaOH)}\) reacted with \(\text{Y}\) and \(\text{Z} = 0.0617 \times 0.02249 = 1.39 \times 10^{-3}\ \text{mol}\)

\(\text{n}_{\text{Y}}= \dfrac{0.100}{72.062}= 0.00139\ \text{mol}\). Therefore \(\text{Y}\) reacts in a \(1:1\) molar ratio as it is a monoprotic acid.

\(\text{n}_{\text{Z}}= \dfrac{0.100}{144.124}= 0.0006938\ \text{mol}\). Therefore \(\text{Z}\) reacts in a \(2:1\) molar ratio with \(\ce{NaOH}\) as it is a diprotic acid.

→ The equal volumes of \(\text{Y}\) and \(\text{Z}\) used in the titration can be attributed to \(\text{Z}\) having twice the molar mass of \(\text{Y}\) and being a diprotic acid.

Show Worked Solution

→ Both sample \(\text{Y}\) and \(\text{Z}\) undergo an addition reaction with bromine and a hydration reaction. Therefore these samples must contain a \(\ce{C=C}\) bond.

→ As sample \(\text{X}\), undergoes neither of these reactions, it must have no \(\ce{C=C}\) bond, thus sample \(\text{X}\) is structure \(2\).

→ Both structure \(1\) and structure \(3\) contain 1 \(\ce{C=C}\) each \(\Rightarrow\) they will react in a \(1:1\) with \(\ce{Br2}\). Hence the same number of moles of the carboxylic acid samples will react with the bromine.

→ Since the mass of sample \(\text{Z}\) that reacts with the bromine is double the mass of sample \(\text{Y}\), the molar mass of sample \(\text{Z}\) must be double the molar mass of sample \(\text{Y}\) following the formula \(m = n \times MM\).

→ Therefore, sample \(\text{Y}\) is structure 1 and sample \(\text{Z}\) is structure 3.
 

Other information provided that could support identification includes:

→ The two products formed for the hydration of \(\text{Y}\) is due to the asymmetry of structure 1 and the single product formed in the hydration of \(\text{Z}\) is due to the symmetrical nature of structure 3.

The titration values are consistent with the proposed samples and their corresponding structures.

→ \(\ce{n(NaOH)}\) reacted with \(\text{X} = 0.0617 \times 0.02188 = 1.35 \times 10^{-3}\ \text{mol}\)

\(\text{n}_{\text{X}}= \dfrac{0.100}{74.078}= 0.00135\ \text{mol}\). Therefore \(\text{X}\) reacts in a \(1:1\) molar ratio as it is a monoprotic acid.

→ \(\ce{n(NaOH)}\) reacted with \(\text{Y}\) and \(\text{Z} = 0.0617 \times 0.02249 = 1.39 \times 10^{-3}\ \text{mol}\)

\(\text{n}_{\text{Y}}= \dfrac{0.100}{72.062}= 0.00139\ \text{mol}\). Therefore \(\text{Y}\) reacts in a \(1:1\) molar ratio as it is a monoprotic acid.

\(\text{n}_{\text{Z}}= \dfrac{0.100}{144.124}= 0.0006938\ \text{mol}\). Therefore \(\text{Z}\) reacts in a \(2:1\) molar ratio with \(\ce{NaOH}\) as it is a diprotic acid.

→ The equal volumes of \(\text{Y}\) and \(\text{Z}\) used in the titration can be attributed to \(\text{Z}\) having twice the molar mass of \(\text{Y}\) and being a diprotic acid.

♦ Mean mark 55%.

CHEMISTRY, M8 2024 HSC 33

Acetone can be reduced, as shown.
 

  1. Identify the shape around the central carbon atom in each molecule.   (2 marks)

  2. Explain how \({ }^{13} \text{C NMR}\) spectroscopy could be used to monitor the progress of this reaction.  (3 marks)

Show Answers Only

a.   Acetone:

→ Double bond and 2 single bonds coming off the central carbon atom  \(\Rightarrow\)  trigonal planar.

Product:

→ Contains single bonds coming off the central carbon atom (\Rightarrow\) tetrahedral. (Note: the hydrogen bonded to the central carbon atom in the product molecule is not shown due to the skeletal structure)
 

b.    \({ }^{13} \text{C NMR}\) Spectroscopy:

→ \({ }^{13} \text{C NMR}\) will differentiate between molecules with different carbon environments. This produces different signals on the \({ }^{13} \text{C NMR}\) spectrum.

→ The acetone would produce two signals on the \({ }^{13} \text{C NMR}\) spectrum. The first signal would be due to the \(\ce{CH3}\) groups either side of the central carbon between 20-50 ppm. The second signal would be from the carbonyl group between 190-220 ppm.

→ The product of the reduction would also produce two signals on the \({ }^{13} \text{C NMR}\) spectrum. The carbon with the hydroxyl group attached to it would produce a signal between 50-90 ppm and the \(\ce{CH3}\) groups either side would produce a signal between 5-40 ppm.

→ The reaction can be monitored by observing the disappearance of the carbonyl signal (190-220 ppm) and appearance of the hydroxyl signal (50-90 ppm) as acetone is reduced to the product.

Show Worked Solution

a.   Acetone:

→ Double bond and 2 single bonds coming off the central carbon atom \(\Rightarrow\) trigonal planar.

Product:

→ Contains single bonds coming off the central carbon atom  \(\Rightarrow\)  tetrahedral. (Note: the hydrogen bonded to the central carbon atom in the product molecule is not shown due to the skeletal structure)

♦ Mean mark (a) 44%.

  

b.    \({ }^{13} \text{C NMR}\) Spectroscopy:

→ \({ }^{13} \text{C NMR}\) will differentiate between molecules with different carbon environments. This produces different signals on the \({ }^{13} \text{C NMR}\) spectrum.

→ The acetone would produce two signals on the \({ }^{13} \text{C NMR}\) spectrum. The first signal would be due to the \(\ce{CH3}\) groups either side of the central carbon between 20-50 ppm. The second signal would be from the carbonyl group between 190-220 ppm.

→ The product of the reduction would also produce two signals on the \({ }^{13} \text{C NMR}\) spectrum. The carbon with the hydroxyl group attached to it would produce a signal between 50-90 ppm and the \(\ce{CH3}\) groups either side would produce a signal between 5-40 ppm.

→ The reaction can be monitored by observing the disappearance of the carbonyl signal (190-220 ppm) and appearance of the hydroxyl signal (50-90 ppm) as acetone is reduced to the product.

CHEMISTRY, M5 2024 HSC 32

Calculate the concentration of cadmium ions in a saturated solution of cadmium(\(\text{II}\)) phosphate,  \(\ce{Cd3\left(PO4\right)2}, \ K_{sp}=2.53 \times 10^{-33}\).   (4 marks)

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\(3.56 \times 10^{-7}\ \text{mol L}^{-1}\)

Show Worked Solution

\(\ce{Cd3(PO4)2(s) \rightleftharpoons 3Cd^{2+}(aq) + 2PO4^{3-}(aq)}\)

\(\Rightarrow K_{sp} = \ce{[Cd^{2+}]^3[PO4^{3-}]^2} = 2.53 \times 10^{-33}\)

→ \(\text{Let the molar solubility of }\ce{Cd3(PO4)2} \text{ be } x\ \text{mol L}^{-1}\)

→ \(\ce{[Cd^{2+}]} = 3x,\ \ \ce{[PO4^{3-}]} = 2x\)

\(\ce{[Cd^{2+}]^3[PO4^{3-}]^2}\) \(=2.53 \times 10^{-33}\)  
\((3x)^3 \times (2x)^2\) \(=2.53 \times 10^{-33}\)  
\(108x^5\) \(=2.53 \times 10^{-33}\)  
\(x\) \(=\sqrt[5]{\dfrac{2.53 \times 10^{-33}}{108}}\)  
  \(=1.11856 \times 10^{-7}\)  

 

→ \(\ce{[Cd^{2+}]} = 3 \times 1.1186 \times 10^{-7} = 3.56 \times 10^{-7}\ \text{mol L}^{-1}\ \ \text{(3 sig.fig)}\)

♦ Mean mark 55%.

CHEMISTRY, M5 2024 HSC 30

An equilibrium mixture of hydrogen, carbon dioxide, water and carbon monoxide is in a closed, 1 L container at a fixed temperature as shown:

\(\ce{H2(g) +CO_2(g) \rightleftharpoons H2O(g) +CO(g)} \quad \quad K_{eq}=1.600\)

The initial concentrations are  \(\left[\ce{H2}\right]=1.000 \text{ mol L}^{-1}, \left[ \ce{CO2}\right]=0.500\ \text{mol L}^{-1},\ \left[\ce{H2O}\right]=0.400 \text{ mol L}^{-1}\)  and  \([ \ce{CO} ]=2.000 \text{ mol L} ^{-1}\).

An unknown amount of \(\ce{CO(g)}\) was added to the same container, and the temperature was kept constant. After the new equilibrium had been established, the concentration of \(\ce{H2O(g)}\) was found to be 0.200 mol L\(^{-1}\).

Using this information, calculate the unknown amount (in mol) of \(\ce{CO(g)}\) that was added to the container.   (4 marks)

Show Answers Only

\(4.92\ \text{mol}\)

Show Worked Solution

→ \(\ce{n_{intial}(CO(g))} = 0.400 \text{ and } \ce{n_{final}(CO(g))} = 0.200\).

→ \(\text{Change in the number of moles in } \ce{CO(g)} = 0.400-0.200 = 0.200\ \text{mol in } 1\ \text{L}\)

\begin{array} {|c|c|c|c|c|}
\hline & \ce{H2(g)} & \ce{CO2(g)} & \ce{H2O(g)} & \ce{CO2(g)} \\
\hline \text{Initial} & 1 & 0.5 & 0.4 & 2 + x \\
\hline \text{Change} & +0.2 & +0.2 & -0.2 & -0.2 \\
\hline \text{Equilibrium} & 1.2 & 0.7 & 0.2 & 1.8 + x \\
\hline \end{array}

→ Since all substances are present in a 1 L container, the concentrations of each substance is equal to the number of moles of that substance present at equilibrium

\(K_{eq}\) \(=\dfrac{\ce{[H2O(g)][CO(g)]}}{\ce{[H2(g)][CO2(g)]}}\)  
\(1.600\) \(=\dfrac{0.2 \times (1.8+x)}{1.2 \times 0.7}\)  
\(1.8 + x\) \(=1.6 \times \dfrac{1.2 \times 0.7}{0.2}\)  
\(x\) \(=6.72-1.8\)  
  \(=4.92\ \text{mol}\)  

 

→ 4.92 mol of \(\ce{CO(g)}\) were added to the container.

♦ Mean mark 55%.

CHEMISTRY, M5 2024 HSC 26

The equilibrium equation for the reaction of iodine with hydrogen cyanide in aqueous solution is given.

\(\ce{I_2(aq) + HCN(aq)\rightleftharpoons ICN(aq) + I^{-}(aq) + H^{+}(aq)}\)

At  \(t=0\) min, \(\ce{I2}\) was added to a mixture of \(\ce{HCN, I^{-}}\) and \(\ce{H^{+}}\), bringing \(\left[ \ce{I2}\right]\) to 2.0 × 10\(^{-5}\) mol L\(^{-1}\). After 3 minutes, the system was at equilibrium, and an analysis of the mixture found that half of the \(\ce{I2}\) had reacted.

  1. On the axes provided, sketch a graph to show how \(\left[\ce{I_2}\right]\) changes in the solution between  \(t=0\) min  and  \(t=6\) min.   (2 marks)

     

  1. Using collision theory, explain the rate of reaction between  \(t=0\) min  and  \(t=6\) min. Refer to the \(\left[ \ce{I2}\right]\) in your answer.   (3 marks)

Show Answers Only

a.    

b.    Initially, the high \(\ce{[I2]}\) results in a large number of collisions between reactants.

→ This results in an initially high reaction rate for the forward reaction. 

→ Between \(t=0\) and \(t=3\), the concentration of \(\ce{I2}\) decreases. Less collisions between reactant particles occur leading to a decrease in the rate of the forward reaction.

→ Between \(t=0\) and \(t=3\), as the concentration of reactants decrease, the concentration of the products increase. This leads to an increase in the number of successful collisions between product particles and a subsequent increase in the rate of the reverse reaction.

→ Between \(t=3\) and \(t=6\), the concentration of \(\ce{I2}\) remains constant. This is due to the system reaching a state of dynamic equilibrium so the frequency of successful collisions between the reactants is equal to the frequency of successful collisions between the products. i.e. the rates of the forward and reverse reactions are equal. 

Show Worked Solution

a.   
     
 

b.    Initially, the high \(\ce{[I2]}\) results in a large number of collisions between reactants.

→ This results in an initially high reaction rate for the forward reaction. 

→ Between \(t=0\) and \(t=3\), the concentration of \(\ce{I2}\) decreases. Less collisions between reactant particles occur leading to a decrease in the rate of the forward reaction.

→ Between \(t=0\) and \(t=3\), as the concentration of reactants decrease, the concentration of the products increase. This leads to an increase in the number of successful collisions between product particles and a subsequent increase in the rate of the reverse reaction.

→ Between \(t=3\) and \(t=6\), the concentration of \(\ce{I2}\) remains constant. This is due to the system reaching a state of dynamic equilibrium so the frequency of successful collisions between the reactants is equal to the frequency of successful collisions between the products. i.e. the rates of the forward and reverse reactions are equal. 

Mean mark (b) 56%.

CHEMISTRY, M6 2024 HSC 20 MC

The concentration of ascorbic acid \(\left(M M=176.124\ \text{g mol}^{-1}\right)\) in solution \(\text{A}\) was determined by titration.

    • A 25.00 mL sample of solution \(\text{A}\) was titrated with potassium hydroxide solution.
    • 50.00 mg of ascorbic acid was added to a second 25.00 mL sample of solution \(\text{A}\), which was titrated in the same way.

Titration volumes for both titrations are given.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textit{Solution} \rule[-1ex]{0pt}{0pt} & \textit{Titre}\text{ (mL)} \\
\hline
\rule{0pt}{2.5ex} \text{25.00 mL solution A} \rule[-1ex]{0pt}{0pt} & 17.50 \\
\hline
\rule{0pt}{2.5ex} \text{25.00 mL solution A+}& 33.10 \\
\text{50.00 mg of ascorbic acid}& \text{} \\
\hline
\end{array}

What is the concentration of ascorbic acid in solution \(\text{A}\)?

  1. \(5.352 \times 10^{-3} \text{ mol L}^{-1}\)
  2. \(6.004 \times 10^{-3} \text{ mol L}^{-1}\)
  3. \(1.012 \times 10^{-2} \text{ mol L}^{-1}\)
  4. \(1.274 \times 10^{-2} \text{ mol L}^{-1}\)
Show Answers Only

\(D\)

Show Worked Solution

→ \(\ce{n(ascorbic acid)} = \dfrac{0.05}{176.124} = 0.00028389\ \text{mol}\)

→ Volume of \(\ce{KOH}\) to neutralise the two samples = 15.6 mL.

→ 15.6 mL was required to neutralise the extra 0.00028389 mol of ascorbic acid.

→ Concentration of potassium hydroxide \(=\dfrac{\text{n}}{\text{V}}= \dfrac{0.00028389}{0.0156} = 0.018198\ \text{mol L}^{-1}\)

→ \(\ce{n(KOH)}\) used in the first titration \(\ce{= c \times V} = 0.018198 \times 0.0175 = 0.0003185\ \text{mol} = \ce{n(ascorbic acid)}\)

→ \(\ce{[ascrobic acid]} = \dfrac{0.0003185}{0.0250} = 0.01274 = 1.274 \times 10^{-2}\ \text{mol L}^{-1}\)

\(\Rightarrow D\)

♦ Mean mark 43%.

CHEMISTRY, M8 2024 HSC 19 MC

Which of the following compounds produces TWO doublets in the \({ }^1 \text{H NMR}\) spectrum?
 


 

Show Answers Only

\(A\)

Show Worked Solution

→ For two doublets to occur, the substance must have two chemically different hydrogen environments that both have one neighbouring hydrogen on an adjacent carbon atom.
 

→ This is displayed by option \(\text{A}\) where the hydrogen atoms in environments 1 and 3 both have only one neighbouring hydrogen atom each. 

\(\Rightarrow A\)

♦ Mean mark 48%.

CHEMISTRY, M5 2024 HSC 18 MC

A reaction mixture, not at equilibrium, is composed of both \(\ce{N_2O_4(g)}\) and \(\ce{NO_2(g)}\) in a closed container. The reaction quotient for the system, \(Q\), is given.

\(Q=\dfrac{\left[\ce{NO_2}\right]^2}{\left[\ce{N_2O_4}\right]}\)

The rate of the forward reaction is initially greater than the rate of the reverse reaction.

Which diagram shows how \(Q\) changes over time for this mixture?
 

Show Answers Only

\(B\)

Show Worked Solution

→ When calculating the reaction quotient of a chemical reaction, the formula is \(\dfrac{products}{reactants}\).

→ The equation for the reaction taking place is:

\(\ce{N2O4(g) \rightleftharpoons 2NO2(g)}\)

→ As the rate of the forward reaction is greater than the rate of the reverse reaction, \(\ce{[NO2]^2}\) will increase and \(\ce{[N2O4]}\) will decrease.

→ Hence the value for the reaction quotient, \(\dfrac{\left[\ce{NO_2}\right]^2}{\left[\ce{N_2O_4}\right]}\), will increase.

→ As it states both \(\ce{N2O4(g)}\) and \(\ce{NO2(g)}\) are present in the intial system, the value for \(Q\) will not be zero.

\(\Rightarrow B\)

♦ Mean mark 43%.

CHEMISTRY, M6 2024 HSC 17 MC

20 mL of a 0.1 mol L\(^{-1}\) solution of an acid is titrated against a 0.1 mol L\(^{-1}\) solution of sodium hydroxide. A graph of pH against the volume of sodium hydroxide for this experiment is shown.
 

   

Which of the following acids was used in the titration?

\begin{align*}
\begin{array}{l}
\ & \\
\textbf{A.}\\
\textbf{B.}\\
\textbf{C.}\\
\textbf{D.}\\
\end{array}
\begin{array}{|c|c|c|}
\hline
\quad \textit{Acid}\quad  & \quad pK_{a1}\quad & \quad pK_{a2}\quad \\
\hline
1& 4.76 & – \\
\hline
2 & \text{Strong} & – \\
\hline
3 & 1.91 & 6.30 \\
\hline
4 & 4.11 & 9.61 \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(C\)

Show Worked Solution

→ There are two equivalence points on the graph, so the acid is diprotic and will have two \(pK_a\) values.

→ The reaction at the first equivalence point is: 

\(\ce{H2A -> HA^- + H^+}\)

→ The reaction at the second equivalence point is: 

\(\ce{HA^- -> A^{2-} + H^+}\)

→ The \(pK_a\) values for each equivalence point will be equal to the pH values at the half-equivalence points.

→ The first half-equivalence point occurs at 10 mL, where the corresponding pH is 1.91 \(\Rightarrow pK_{a1}\) = 1.91

→ The second half-equivalence point will occur halfway between the first two equivalence points, at 30 mL

→ The corresponding pH is 6.30 \(\Rightarrow pK_{a2}\) = 6.30

\(\Rightarrow C\)

♦ Mean mark 41%.
COMMENT: Student’s did well understanding the relationship between the \(pK_a\) values and half-equivalence points.

CHEMISTRY, M5 2024 HSC 15 MC

The thermal decomposition of lithium peroxide \(\ce{(Li_2O_2)}\) is given by the equation shown.

\(\ce{2Li_2O_2(s)\rightleftharpoons 2 Li_2 O(s) + O_2(g)}\)

Mixtures of \(\ce{Li_2O_2}\), \(\ce{Li_2O}\) and \(\ce{O_2}\) were allowed to reach equilibrium in two identical, closed containers, \(\text{P}\) and \(\text{Q}\), at the same temperature. The amount of \(\ce{Li_2O_2(s)}\) in container \(\text{P}\) is double that in container \(\text{Q}\) . The amount of \(\ce{Li_2O(s)}\) is the same in each container.

What is the ratio of \(\left[ \ce{O_2(g)}\right]\) in container \(\text{P}\) to \(\left[\ce{O_2(g)}\right]\) in container \(\text{Q}\)?

  1. \(1: 1\)
  2. \(2: 1\)
  3. \(3: 2\)
  4. \(5: 4\)
Show Answers Only

\(A\)

Show Worked Solution

→ When calculating the \(K_{eq}\) of a system, substances in solid states are all given a value of \(1\).

→ The equilibrium constant of the above reaction is \(K_{eq} = \ce{[O_2(g)]}\).

→ As both mixtures reached equilibrium, the \(K_{eq}\) values for each mixture is the same, hence the ratio of \(\ce{[O2(g)]}\) in each container is \(1:1\).

\(\Rightarrow A\)

♦♦ Mean mark 39%.

CHEMISTRY, M5 2024 HSC 11 MC

Which is the correct expression for calculating the solubility (in mol L\(^{-1}\)) of lead\(\text{(II}\)) iodide in a 0.1 mol L\(^{-1}\) solution of \(\ce{NaI}\) at 25°C?

  1. \(\dfrac{9.8 \times 10^{-9}}{2 \times 0.1}\)
  2. \(\dfrac{9.8 \times 10^{-9}}{(2 \times 0.1)^2}\)
  3. \(\dfrac{9.8 \times 10^{-9}}{0.1}\)
  4. \(\dfrac{9.8 \times 10^{-9}}{(0.1)^2}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\ce{PbI2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)}\)

→ Let the solubility of \(\ce{PbI2} = x\ \text{mol L}^{-1}\)

\begin{array} {|c|c|c|c|}
\hline  \text{Concentration (mol/L)} & \ce{PbI2} & \ce{Pb^{2+}} & \ce{2I^-} \\
\hline \text{Initial} & – & 0 & 0.1 \\
\hline \text{Change} & – & +x & +2x \\
\hline \text{Equilibrium} & – & x & 0.1 +2x \\
\hline \end{array}

→ As \(x\) is very small, assume  \(0.1 +2x \approx 0.1\)

\(K_{sp} \ce{=[Pb^{2+}][I^-]^2} =9.8 \times 10^{-9}\)

\(K_{sp} = 0.1^2 \times x =9.8 \times 10^{-9}\)

\(\therefore x=\dfrac{9.8 \times 10^{-9}}{0.1^2}\)

\(\Rightarrow D\)

♦ Mean mark 55%.

Matrices, GEN2 2024 VCAA 12

When the construction company established the construction site at the beginning of 2023, it employed 390 staff to work on the site.

The staff comprised 330 construction workers \((C)\), 50 foremen \((F)\) and 10 managers \((M)\).

At the beginning of each year, staff can choose to stay in the same job, move to a different job on the site, or leave the site \((L)\) and not return.

The transition diagram below shows the proportion of staff who are expected to change their job at the site each year.
 

This situation can be modelled by the recurrence relation

\(S_{n+1}=T S_n\), where

\(T\) is the transitional matrix, \(S_0=\left[\begin{array}{c}330 \\ 50 \\ 10 \\ 0\end{array}\right] \begin{aligned} & C \\ & F \\ & M \\ & L \end{aligned}\)  and \(n\) is the number of years after 2023.

  1. Calculate the predicted percentage decrease in the number of foremen \((F)\) on the site from 2023 to 2025.   (1 mark)

  2. Determine the total number of staff on the site in the long term.  (1 mark)

To encourage more construction workers \((C)\) to stay, the construction company has given workers an incentive to move into the job of foreman \((F)\).

Matrix \(R\) below shows the ways in which staff are expected to change their jobs from year to year with this new incentive in place.

\begin{aligned}
& \quad \quad \ \ \textit{this year} \\
& \quad  C \quad  \ \ F \quad  \ \  M \quad L\\
R = & \begin{bmatrix}
0.4 & 0.2 & 0 & 0 \\
0.4 & 0.2 & 0.4 & 0 \\
0 & 0.2 & 0.3 & 0 \\
0.2 & 0.4 & 0.3 & 1
\end{bmatrix}\begin{array}{l}
C\\
F\\
M\\
L
\end{array} \quad \textit{next year}
\end{aligned}

The site always requires at least 330 construction workers.

To ensure that this happens, the company hires an additional 190 construction workers \((C)\) at the beginning of 2024 and each year thereafter.

The matrix  \(V_{n+1}\)  will then be given by

\(V_{n+1}=R V_n+Z\), where

\(V_0=\left[\begin{array}{c}330 \\ 50 \\ 10 \\ 0\end{array}\right] \begin{aligned} & C \\ & F \\ & M \\ & L\end{aligned} \quad\quad\quad Z=\left[\begin{array}{c}190 \\ 0 \\ 0 \\ 0\end{array}\right] \begin{aligned} & C \\ & F \\ & M \\ & L\end{aligned} \ \ \) and \(n\) is the number of years after 2023.

  1. How many more staff are there on the site in 2024 than there were in 2023 ?    (1 mark)

  2. Based on this new model, the company has realised that in the long term there will be more than 200 foremen on site.
  3. In which year will the number of foremen first be above 200?   (1 mark)

Show Answers Only

a.    \(14\%\)

b.    \(\text{Zero}\)

c.    \(101\)

d.    \(2027\)

Show Worked Solution

a.    \(\text{Using CAS:}\)

\(\text{Transition matrix}(T):\ \begin{aligned}
& \quad \quad \quad \ \ \ \ \textit{from} \\
& \quad \ \ \ \  C \ \ \ \ \  F \ \ \ \ \ \  M \ \ \ \ \ \  L \ \ \\
\ \ \textit{to}\ \ \ & \begin{array}{l}
C \\
F \\
M \\
L
\end{array}\begin{bmatrix}
0.3 & 0.2 & 0 & 0 \\
0.2 & 0.2 & 0.4 & 0 \\
0 & 0.2 & 0.3 & 0 \\
0.5 & 0.4 & 0.3 & 1
\end{bmatrix}
\end{aligned}\)

♦♦♦ Mean mark (a) 25%.

\(S_{2023}=\begin{bmatrix}
330 \\
50 \\
10 \\
0 \\
\end{bmatrix}\ \ ,\ \  S_{2024}=T\times S_{2023}=\begin{bmatrix}
109 \\
80 \\
13 \\
188 \\
\end{bmatrix}\)

\( S_{2025}=T\times S_{2024}=\begin{bmatrix}
48.7 \\
43 \\
19.9 \\
278.4 \\
\end{bmatrix}\)

 
\(\text{% decrease in foremen}\ (F)=\dfrac{50-43}{50}\times 100\%=14\%\)
 

b.   \(\text{Test for 10 years:}\)

\(\text{Using CAS:}\)

\(S_{2024}=T^{10}\times S_{2023}=\begin{bmatrix}
0.490748 \\
0.734232 \\
0.488858 \\
388.286 \\
\end{bmatrix}\)

 
\(\text{In the long term, there will be zero employees on site.}\)
 

♦♦♦ Mean mark (b) 24%.

c.    \(\text{Using CAS:}\)

\(V_{2024}=R\times V_{2023}+Z=\begin{bmatrix}
332 \\
146 \\
13 \\
89 \\
\end{bmatrix}\)
  

\(\text{Difference in staff from}\ 2023-2024\)

\(=(332+146+13)-390\)

\(=101\ \text{more staff.}\)
 

\(\text{NOTE: 89 not included in calculation as these are the staff}\)

\(\text{who have left the company during the year.}\)
 

♦♦♦ Mean mark (c) 12%.

d.   \(\text{Using CAS:}\)

\(V_{2024}=R\times V_{2023}+Z=\begin{bmatrix}
332 \\
146 \\
13 \\
89 \\
\end{bmatrix}\ \ ,\ \  V_{2025}=R\times V_{2024}+Z=\begin{bmatrix}
352 \\
167.2 \\
33.1 \\
217.7 \\
\end{bmatrix}\ \\\)

♦♦♦ Mean mark (d) 26%.

\(V_{2026}=R\times V_{2025}+Z=\begin{bmatrix}
364.24\\
187.48 \\
43.37 \\
364.91 \\
\end{bmatrix}\ \ ,\ \  V_{2027}=R\times V_{2026}+Z=\begin{bmatrix}
373.192 \\
200.54 \\
50.507 \\
525.761 \\
\end{bmatrix}\ \\\)

 
\(\therefore\ \text{In 2027 the number of foremen will be over 200.}\)

Matrices, GEN2 2024 VCAA 11

A population of a native animal species lives near the construction site.

To ensure that the species is protected, information about the initial female population was collected at the beginning of 2023. The birth rates and the survival rates of the females in this population were also recorded.

This species has a life span of 4 years and the information collected has been categorised into four age groups: 0-1 year, 1-2 years, 2-3 years, and 3-4 years.

This information is displayed in the initial population matrix, \(R_0\), and the Leslie matrix, \(L\), below.

\(R_0=\left[\begin{array}{c}70 \\ 80 \\ 90 \\ 40\end{array}\right] \quad \quad L=\left[\begin{array}{cccc}0.4 & 0.75 & 0.4 & 0 \\ 0.4 & 0 & 0 & 0 \\ 0 & 0.7 & 0 & 0 \\ 0 & 0 & 0.5 & 0\end{array}\right]\)

  1. Using the information above
  2.  i. complete the following transition diagram.   (1 mark) 

  1. ii. complete the following table, showing the initial female population, and the predicted female population after one year, for each of the age groups.  (1 mark)  

  1. It is predicted that if this species is not protected, the female population of each of the four age groups will rapidly decrease within the next 10 years.
  2. After how many years is it predicted that the total female population of this species will first be half the initial female population?   (1 mark)

Show Answers Only

a.i. 

a.ii.

  \(\textbf{Age Group}\)
  \(\ 0-1\ \text{year}\ \) \(\ 1-2\ \text{years}\ \) \(\ 2-3\ \text{years}\ \) \(\ 3-4\ \text{years}\ \)
\(\ \text{Initial population}\ \) \(70\) \(80\) \(90\) \(40\)
\(\ \text{Population after}\ \)
\(\ \text{one year}\)		 \(124\) \(28\) \(56\) \(45\)

 

b.    \(\text{5 years}\)

Show Worked Solution

a.i. 

♦♦♦ Mean mark (a) 24%.

a.ii.  \(\text{Population after 1 yr calculations}\)

\(0-1\ \text{year}\ =0.4\times 70+0.75\times 80+0.4\times 90=124\)

\(1-2\ \text{years}\ =0.4\times 70=28\)

\(2-3\ \text{years}\ =0.7\times 80=56\)

\(3-4\ \text{years}\ =0.5\times 90=45\)

 

  \(\textbf{Age Group}\)
  \(\ 0-1\ \text{year}\ \) \(\ 1-2\ \text{years}\ \) \(\ 2-3\ \text{years}\ \) \(\ 3-4\ \text{years}\ \)
\(\ \text{Initial population}\ \) \(70\) \(80\) \(90\) \(40\)
\(\ \text{Population after}\ \)
\(\ \text{one year}\)		 \(124\) \(28\) \(56\) \(45\)

  

b.    \(\text{Using CAS:}\)

\(R_1=L\times R_0=\begin{bmatrix}
124  \\
28 \\
56  \\
45  \end{bmatrix}\ \ \text{Total = 253}\ ,\ \ R_2=L\times R_1=\begin{bmatrix}
93  \\
49.6 \\
19.6  \\
28  \end{bmatrix}\ \ \text{Total = 190.2}\)

 

\(R_3=L\times R_2=\begin{bmatrix}
82.24  \\
37.2 \\
34.72  \\
9.8  \end{bmatrix}\ \ \text{Total = 163.96}\ ,\ \ R_4=L\times R_3=\begin{bmatrix}
74.684  \\
32.896 \\
26.04  \\
17.36  \end{bmatrix}\ \ \text{Total =150.98}\)

 

\(R_5=L\times R_4=\begin{bmatrix}
64.9616 \\
29.8736 \\
23.0272 \\
13.02 \end{bmatrix}\ \ \text{Total = 130.8824}\)

 
\(\therefore\ \text{Total female population less than 140 after 5 years}\)

♦ Mean mark (b) 39%.

Matrices, GEN2 2024 VCAA 10

To access the southern end of the construction site, Vince must enter a security code consisting of five numbers.

The security code is represented by the row matrix \(W\).

The element in row \(i\) and column \(j\) of \(W\) is \(w_{i j}\).

The elements of \(W\) are determined by the rule  \((i-j)^2+2 j\).

  1. Complete the following matrix showing the five numbers in the security code.   (1 mark)

To access the northern end of the construction site, Vince enters a different security code, consisting of eight numbers.

This security code is represented by the row matrix \(X\).

The element in row \(i\) and column \(j\) of \(X\) is \(x_{i j}\).

The elements of \(X\) are also determined by the rule  \((i-j)^2+2 j\).

  1. What is the last number in this security code to access the northern end of the construction site?  (1 mark)

Show Answers Only

a.    \(W=\ [2\quad 5\quad 10\quad 17\quad 26\ ]\)

b.    \(x_{18}=65\)

Show Worked Solution

a.     \(W\) \(=\ [(0)^2+2\ \ \  (-1)^2+4\ \ \  (-2)^2+36\ \ \  (-3)^2+8\ \ \  (-4)^2+10\ ]\)
    \(=\ [\ 2\quad 5\quad 10\quad 17\quad 26\ ]\)

 

b.    \(x_{18}=(1-8)^2+2\times 8=65\)

Mean mark (a) 54%.
♦ Mean mark (b) 50%

Recursion and Finance, GEN2 2024 VCAA 8

Emi takes out a reducing balance loan of $500 000.

The interest rate is 5.3% per annum, compounding monthly.

Emi makes regular monthly repayments of $3071.63 for the duration of the loan, with only the final repayment amount being slightly different from all the other repayments.

Determine the total cost of Emi's loan, rounding your answer to the nearest cent, and state the number of payments required to fully repay the loan.   (2 marks)
 


Show Answers Only

\(\text{Total cost}\ =$884\,633.62\ ,\ 288\ \text{payments}\)

Show Worked Solution

\(\text{Using CAS: Find}\ N\ \text{when}\ FV=0\)

♦♦ Mean mark 29%.

\(\text{Using CAS: Find amount still owing if 288 payments are made.}\)

\(\text{Final payment}\ =$3071.63+$4.18 =$3075.81\)

\(\text{Total cost of loan}\ =3071.63\times 287 + 3075.81=$884\,633.62\)

\(\text{Total number of payments}\ =288\)

Recursion and Finance, GEN2 2024 VCAA 7

Emi decides to invest a $300 000 inheritance into an annuity.

Let \(E_n\) be the balance of Emi's annuity after \(n\) months.

A recurrence relation that can model the value of this balance from month to month is

\(E_0=300\,000, \quad E_{n+1}=1.003 E_n-2159.41\)

  1. Showing recursive calculations, determine the balance of the annuity after two months. Round your answer to the nearest cent.   (1 mark)

  2. For how many years will Emi receive the regular payment?  (1 mark)

  3. Calculate the annual compound interest rate for this annuity.  (1 mark)

  4. If Emi wanted the annuity to act as a perpetuity, what monthly payment, in dollars, would she receive?  (1 mark)

Show Answers Only

a.    \($297\,477.40\) 

b.    \(15\ \text{years}\)

c.    \(3.6\%\)

d.    \($900\)

Show Worked Solution

a.   \(E_0=300\,000\)

\(E_1=1.003\times 300\,000-2159.41=$298\,740.59\)

\(E_2=1.003\times 298\,740.59-2159.41=$297\,477.4018\approx $297\,477.40\)

♦ Mean mark (a) 49%.

b.    \(\text{Using CAS:}\)

\(\text{Number of years}\ =\dfrac{180}{12}=15\ \text{years}\)
 

♦ Mean mark (b) 42%.

c.    \(\text{Annual interest rate}\ =(1.003-1)\times 12\times 100\% = 3.6\%\)
 

d.   \(\text{Method 1: Using CAS}\)

\(\text{Monthly payment}\ =$900\)
  

\(\text{Method 2:}\)

\(\text{Amount must be equal to the amount of monthly interest earned.}\)

\(\therefore\ \text{Monthly payment}\ =300\,000\times 0.003=$900\)

♦ Mean mark (d) 47%.

Vectors, EXT2 V1 2024 HSC 15a

Consider the three vectors  \(\underset{\sim}{a}=\overrightarrow{O A}, \underset{\sim}{b}=\overrightarrow{O B}\) and \(\underset{\sim}{c}=\overrightarrow{O C}\), where \(O\) is the origin and the points \(A, B\) and \(C\) are all different from each other and the origin.

The point \(M\) is the point such that  \(\dfrac{1}{2}(\underset{\sim}{a}+\underset{\sim}{b})=\overrightarrow{O M}\).

  1. Show that \(M\) lies on the line passing through \(A\) and \(B\).   (1 mark)

  2. The point \(G\) is the point such that  \(\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})=\overrightarrow{O G}\).
  3. Show that \(G\) lies on the line passing through \(M\) and \(C\), and lies between \(M\) and \(C\).   (2 marks)

  4. The complex numbers \(x, w\) and \(z\) are all different and all have modulus 1.
  5. Using part (ii), or otherwise, show that  \(\dfrac{1}{3}(x+w+z)\) is never a cube root of \(x w z\).   (2 marks)

Show Answers Only

i.    \(\text{Equation of line through \(A\) and \(B\)}\)

\(\Rightarrow \ell_1=\overrightarrow{O A}+\lambda \overrightarrow{A B}\)

  \(\overrightarrow{O M}\) \(=\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\underset{\sim}{a}-\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\overrightarrow{O A}+\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})\)
    \(=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{A B}\)

 
\(\therefore \overrightarrow{OM} \ \text{lies on} \ \ell_1\).
 

ii.    \(\text{Equation of line through \(M\) and \(C\)}\)

\(\Rightarrow \ell_2=\overrightarrow{OC}+\lambda \overrightarrow{CM}\)

  \(\overrightarrow{O G}\) \(=\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})\)
    \(=\underset{\sim}{c}-\dfrac{2}{3} \underset{\sim}{c}+\dfrac{1}{3} \underset{\sim}{a}+\dfrac{1}{3} \underset{\sim}{b}\)
    \(=\overrightarrow{OC}+\dfrac{2}{3}\left(\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}-\underset{\sim}{c}\right)\)
    \(=\overrightarrow{OC}+\dfrac{2}{3} \overrightarrow{CM}\)

 
\(\therefore \overrightarrow{O G} \ \ \text{lies on} \ \ \ell_2\)

\(\ \ \overrightarrow{O G} \neq \overrightarrow{O C}  \ \ \text{and} \ \ \overrightarrow{O G} \neq \overrightarrow{O M}\)

\(\therefore G \ \ \text{lies between} \ \ C \ \text{and} \ M\).
 

iii.  \(\text{Place}\ x, w,\ \text{and}\ z\ \text{on unit circle.}\)
 

\(\abs{w}=\abs{x}=\abs{z}=1\)

\(\text{Using part (ii):}\)

\(G \equiv \dfrac{1}{3}(x+w+z)\)

\(G \ \text{lies on} \ CM \Rightarrow G \ \text{is inside the unit circle.}\)

\(\Rightarrow\left|\dfrac{1}{3}(x+w+z)\right|<1\)

\(\text{Since}\ \ \abs{xwz}=\abs{x}\abs{w}\abs{z}=1\)

\(\Rightarrow \ \text{All cube roots have modulus = 1.}\)

\(\therefore \dfrac{1}{3}(x+w+z) \ \ \text{cannot be a cube root of  \(xwz\).}\)

Show Worked Solution

i.    \(\text{Equation of line through \(A\) and \(B\)}\)

\(\Rightarrow \ell_1=\overrightarrow{O A}+\lambda \overrightarrow{A B}\)

  \(\overrightarrow{O M}\) \(=\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\underset{\sim}{a}-\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\overrightarrow{O A}+\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})\)
    \(=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{A B}\)

 
\(\therefore \overrightarrow{OM} \ \text{lies on} \ \ell_1\).
 

ii.    \(\text{Equation of line through \(M\) and \(C\)}\)

\(\Rightarrow \ell_2=\overrightarrow{OC}+\lambda \overrightarrow{CM}\)

  \(\overrightarrow{O G}\) \(=\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})\)
    \(=\underset{\sim}{c}-\dfrac{2}{3} \underset{\sim}{c}+\dfrac{1}{3} \underset{\sim}{a}+\dfrac{1}{3} \underset{\sim}{b}\)
    \(=\overrightarrow{OC}+\dfrac{2}{3}\left(\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}-\underset{\sim}{c}\right)\)
    \(=\overrightarrow{OC}+\dfrac{2}{3} \overrightarrow{CM}\)

 
\(\therefore \overrightarrow{O G} \ \ \text{lies on} \ \ \ell_2\)

\(\ \ \overrightarrow{O G} \neq \overrightarrow{O C}  \ \ \text{and} \ \ \overrightarrow{O G} \neq \overrightarrow{O M}\)

\(\therefore G \ \ \text{lies between} \ \ C \ \text{and} \ M\).

♦ Mean mark (ii) 43%.

iii.  \(\text{Place}\ x, w,\ \text{and}\ z\ \text{on unit circle.}\)
 

♦♦♦ Mean mark (iii) 10%.

\(\abs{w}=\abs{x}=\abs{z}=1\)

\(\text{Using part (ii):}\)

\(G \equiv \dfrac{1}{3}(x+w+z)\)

\(G \ \text{lies on} \ CM \Rightarrow G \ \text{is inside the unit circle.}\)

\(\Rightarrow\left|\dfrac{1}{3}(x+w+z)\right|<1\)

\(\text{Since}\ \ \abs{xwz}=\abs{x}\abs{w}\abs{z}=1\)

\(\Rightarrow \ \text{All cube roots have modulus = 1.}\)

\(\therefore \dfrac{1}{3}(x+w+z) \ \ \text{cannot be a cube root of  \(xwz\).}\)

Recursion and Finance, GEN2 2024 VCAA 6

Emi invested profits of $10 000 into a savings account that earns interest compounding fortnightly, for one year.

The effective interest rate, rounded to two decimal places, is 5.07%.

Assume that there are exactly 26 fortnights in a year.

  1. What is the nominal percentage rate of interest for the account?
  2. Round your answer to two decimal places.   (1 mark)

  3. Explain why the nominal interest rate appears lower than the effective interest rate.  (1 mark)

Show Answers Only

a.    \(4.95\%\ \text{(2 d.p.)}\)

b.    \(\text{Nominal and effective interest rates are equal if there is only one}\)

\(\text{compounding period per year. If there are more compounding}\)

\(\text{periods (as in this example) the effective rate will be higher than}\)

\(\text{the nominal rate.}\)

Show Worked Solution

a.    \(\text{Using CAS: FINANCE}\ \rightarrow \ \text{Interest Conversion}\ \rightarrow \ \text{Nominal Interest Rate}\)

\(\text{nom}(5.07, 26) = 4.95037\%\approx 4.95\%\ \text{(2 d.p.)}\)
 

\(\text{Using formula:}\)

Mean mark (a) 51%.
\(r_e\) \(=\left[\left(1+\dfrac{r}{100n}\right)^n-1\right]\times 100\%\)
\(\dfrac{5.07}{100}\) \(=\left[\left(1+\dfrac{r}{2600}\right)^{26}-1\right]\)
\(\left(1+\dfrac{r}{2600}\right)^{26}\) \(=1.0507\)
\(\dfrac{r}{2600}\) \(=(1.0507)^\frac{1}{26}-1\)
\(r\) \(=\left((1.0507)^\frac{1}{26}-1\right)\times 2600\)
  \(=4.95036\dots\approx 4.95\%\ \text{(2 d.p.)}\)

 
b.    \(\text{Nominal and effective interest rates are equal if there is only one}\)

\(\text{compounding period per year. If there are more compounding}\)

\(\text{periods (as in this example) the effective rate will be higher than}\)

\(\text{the nominal rate.}\)

♦♦♦ Mean mark (b) 26%.

Recursion and Finance, GEN2 2024 VCAA 5

Emi operates a mobile dog-grooming business.

The value of her grooming equipment will depreciate.

Based on average usage, a rule for the value, in dollars, of the equipment, \(V_n\), after \(n\) weeks is

\(V_n=15000-60 n\)

Assume that there are exactly 52 weeks in a year.

  1. By what amount, in dollars, does the value of the grooming equipment depreciate each week?   (1 mark)

  2. Emi plans to replace the grooming equipment after four years.   
  3. What will be its value, in dollars, at this time?   (1 mark)

  4. \(V_n\) is the value of the grooming equipment, in dollars, after \(n\) weeks.   
  5. Write a recurrence relation in terms of \(V_0, V_{n+1}\) and \(V_n\) that can model this value from one week to the next.   (1 mark)

  6. The value of the grooming equipment decreases from one year to the next by the same percentage of the original $15 000 value.
  7. What is this annual flat rate percentage?   (1 mark)

Show Answers Only

a.    \($60\)

b.    \($2520\)

c.    \(V_0=15\,000 , \ \ V_{n+1}=V_n-60\)

d.    \(20.8\%\)

Show Worked Solution

a.    \($60\)
 

b.    \(n=4\times 52=208\)

\(V_{208}\) \(=15\,000-60\times208\)
  \(=$2520\)

 
c.    \(V_0=15\,000 , \ \ V_{n+1}=V_n-60\)
 

d.    \(\text{Flat rate}\ =\dfrac{60}{15\,000}\times 52\times 100\%=20.8\%\)

♦ Mean mark (d) 42%.

Data Analysis, GEN2 2024 VCAA 4

The time series plot below shows the gold medal-winning height for the women's high jump, \(\textit{Wgold}\), in metres, for each Olympic year, \(year\), from 1952 to 1988.
 

A five-median smoothing process will be used to smooth the time series plot above.

The first two points have been placed on the graph with crosses (X) and joined by a dashed line (---).

  1. Complete the five-median smoothing by marking smoothed values with crosses (X) joined by a dashed line (---) on the time series plot above.   (1 mark)

  2. Identify two qualitative features that best describe the time series plot above.  (1 mark)

Show Answers Only

a.     

b.    \(\text{Random fluctuations, increasing trend.}\)

Show Worked Solution

a.    \(\text{Medians are:}\)

\(\textbf{1960}: 1.67, 1.76, \colorbox{lightblue}{1.82}, 1.85, 1.90\)

\(\textbf{1964}: 1.76, 1.82, \colorbox{lightblue}{1.85}, 1.90, 1.92\)

\(\textbf{1968}: 1.82, 1.85, \colorbox{lightblue}{1.90}, 1.92, 1.93\)

\(\textbf{1972}: 1.82, 1.90, \colorbox{lightblue}{1.92}, 1.93, 1.97\)

\(\textbf{1976}: 1.82, 1.92, \colorbox{lightblue}{1.93}, 1.97, 2.02\)

\(\textbf{1980}: 1.92, 1.93, \colorbox{lightblue}{1.97}, 2.01, 2.02\)
  

b.   \(\text{Qualitative features:}\)

\(\text{- Random fluctuations}\)

\(\text{- Increasing trend.}\)

♦♦♦ Mean mark (b) 24%.

Data Analysis, GEN2 2024 VCAA 3

The Olympic gold medal-winning height for the women's high jump, \(\textit{Wgold}\), is often lower than the best height achieved in other international women's high jump competitions in that same year.

The table below lists the Olympic year, \(\textit{year}\), the gold medal-winning height, \(\textit{Wgold}\), in metres, and the best height achieved in all international women's high jump competitions in that same year, \(\textit{Wbest}\), in metres, for each Olympic year from 1972 to 2020.

A scatterplot of \(\textit{Wbest}\) versus \(\textit{Wgold}\) for this data is also provided.

When a least squares line is fitted to the scatterplot, the equation is found to be:

\(Wbest =0.300+0.860 \times Wgold\)

The correlation coefficient is 0.9318

  1. Name the response variable in this equation.   (1 mark)

  2. Draw the least squares line on the scatterplot above.  (1 mark)

  3. Determine the value of the coefficient of determination as a percentage.  (1 mark)
  4. Round your answer to one decimal place.
  5. Describe the association between \(\textit{Wbest}\) and \(\textit{Wgold}\) in terms of strength and direction.  (1 mark)

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text { strength } \rule[-1ex]{0pt}{0pt} & \quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\\
\hline
\rule{0pt}{2.5ex}\text { direction } \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}

  1. Referring to the equation of the least squares line, interpret the value of the slope in terms of the variables \(\textit{Wbest}\) and \(\textit{Wgold}\).  (1 mark)

  2. In 1984, the \(\textit{Wbest}\) value was 2.07 m for a \(\textit{Wgold}\) value of 2.02 m .
  3. Show that when this least squares line is fitted to the scatterplot, the residual value for this point is 0.0328.  (2 marks)

  4. The residual plot obtained when the least squares line was fitted to the data is shown below. The residual value from part f is missing from the residual plot.
     

    1. Complete the residual plot by adding the residual value from part f, drawn as a cross ( X ), to the residual plot above.   (1 mark)

    2. In part b, a least squares line was fitted to the scatterplot. Does the residual plot from part g justify this? Briefly explain your answer.  (1 mark)

  1. In 1964, the gold medal-winning height, \(\textit{Wgold}\), was 1.90m . When the least squares line is used to predict \(\textit{Wbest}\), it is found to be 1.934 m .
  2. Explain why this prediction is not likely to be reliable.  (1 mark)

Show Answers Only

a.    \(Wbest\)

b.    

c.    \(86.8\%\)

d.    \(\text{Strong, positive}\)

e.    \(Wbest\ \text{will increase, on average, by 0.86 metres for every metre of increase in}\ Wgold.\)

f.      \(Wbest\) \(=0.300 +0.86\times 2.02\)
    \(=2.0372\)

\(\therefore\ \text{Residual}\ =2.07-2.0372=0.0328\)

g.i.

g.ii.  \(\text{Yes, it is justified as there is no clear pattern, linear or otherwise.}\)

h.    \(\text{This prediction is outside the data range (1972 – 2020 → extrapolation)}\)

\(\text{and therefore cannot be relied upon.}\)

Show Worked Solution

a.    \(Wbest\)

b.    \(\text{Using points:}\ (1.90, 1.934)\ \text{and}\ (2.00, 2.02)\)
 

Mean mark (b) 51%.

c.    \(r=0.9318\ \ \Rightarrow\ \ r^2=0.9318^2=0.8682\dots\)

\(\therefore\ \text{Coefficient of determination} \approx 86.8\%\)
 

d.    \(\text{Strong, positive}\)
 

e.    \(Wbest\ \text{will increase, on average, by 0.86 metres for every metre of increase in}\ Wgold.\)
 

f.      \(Wbest\) \(=0.300 +0.86\times 2.02\)
    \(=2.0372\)

 
\(\therefore\ \text{Residual}\ =2.07-2.0372=0.0328\)

♦ Mean mark (f) 48%.

g.i.

g.ii.  \(\text{Yes, it is justified as there is no clear pattern, linear or otherwise.}\)

♦ Mean mark (g)(i) 47%.
♦ Mean mark (g)(ii) 40%.

h.    \(\text{This prediction is outside the data range (1972–2020 → extrapolation)}\)

\(\text{and therefore cannot be relied upon.}\)

♦ Mean mark (h) 50%.

BIOLOGY, M7 2024 HSC 13 MC

Which of the following identifies plant responses to pathogens?

  1. Increased phagocytosis and programmed cell death
  2. Increased number of stomata and programmed cell death
  3. Production of antihistamines and increased thickness of cell walls
  4. Production of antimicrobial substances and increased thickness of cell walls
Show Answers Only

\(D\)

Show Worked Solution

→ Plants respond to pathogens by producing antimicrobial substances and strengthening their cell walls for enhanced physical defence.

→ The other options either describe animal immune responses (phagocytosis, antihistamines) or incorrect plant responses (increased stomata).

\(\Rightarrow D\)

♦ Mean mark 47%.

BIOLOGY, M5 2024 HSC 6 MC

The diagram shows the karyotypes of a body cell for a male and a female fruit fly.
 

How many chromosomes will the egg of a female fruit fly have?

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

\(B\)

Show Worked Solution

→ In the diagram, the female fruit fly has 8 chromosomes in body cells (2 sex chromosomes + 6 autosomes across pairs I, II, and III)

→ During gamete formation, the female fruit fly’s cells undergo meiosis which halves the total chromosome number.

→ Therefore her eggs will contain half that number, resulting in 4 chromosomes total.

\(\Rightarrow B\)

♦ Mean mark 50%.

BIOLOGY, M7 2024 HSC 8 MC

Trypanosomes (Trypanosoma brucei) are protozoans that cause African sleeping sickness in humans. The diagram shows the way that the disease is transmitted to humans.
 

Which row of the table identifies the pathogen, vector and method of disease transmission to humans?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex}\ & \\
\ & \rule[-1ex]{0pt}{0pt} \\ 
\rule{0pt}{2.5ex} \textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex} \textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex} \textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex} \textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \textit{Pathogen} & \ \ \textit{Vector} & \ \ \textit{Method of disease} \\
\textit{} \rule[-1ex]{0pt}{0pt} & \textit{} & \ \ \textit{transmission} \\
\hline
\rule{0pt}{2.5ex}\text{Trypanosomes} \rule[-1ex]{0pt}{0pt}&  \text{Tsetse fly}& \text{Direct} \\
\hline
\rule{0pt}{2.5ex}\text{Tsetse fly} \rule[-1ex]{0pt}{0pt}& \text{Cow} & \text{Direct} \\
\hline
\rule{0pt}{2.5ex}\text{Trypanosomes} \rule[-1ex]{0pt}{0pt}& \text{Tsetse fly} & \text{Indirect}\\
\hline
\rule{0pt}{2.5ex}\text{Tsetse fly} \rule[-1ex]{0pt}{0pt}&  \text{Cow} & \text{Indirect}\\
\hline
\end{array}
\end{align*}

Show Answers Only

\(C\)

Show Worked Solution

→ African sleeping sickness is caused by trypanosome parasites (pathogen).

→ The pathogens are transmitted indirectly to humans through the bite of tsetse flies (vector).

\(\Rightarrow C\)

♦ Mean mark 48%.

Networks, GEN1 2024 VCAA 40 MC

A project has 15 activities, \(A-O\), that need to be completed.

The directed network that represents this project is shown below.

The activities are not labelled.
 

 

The activity table that could represent this project is
 

 

Show Answers Only

\(B\)

Show Worked Solution

\(\text{A focus are would be the two activities with 4 immediate predecessors.}\)

\(\text{Using trial and error:}\)

♦♦ Mean mark 32%.


  

\(\Rightarrow B\)

Matrices, GEN1 2024 VCAA 37 MC

The network below represents paths through a park from the carpark to a lookout.

The vertices represent various attractions, and the numbers on the edges represent the distances between them in metres.
 

The shortest path from the carpark to the lookout is 34 m .

This can be achieved when

  1. \(x=8\)  and  \(y=8\)
  2. \(x=9\)  and  \(y=7\)
  3. \(x=10\)  and  \(y=6\)
  4. \(x=11\)  and  \(y=5\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{By trial and error}\)

\(\text{Consider Option D:}\ \ \ x=11, y=5\)

\(\text{Shortest path}\ = 10+11+8+5=34\ \checkmark\)
  

 
\(\Rightarrow D\)


♦ Mean mark 40%

Matrices, GEN1 2024 VCAA 31 MC

The matrix below shows the results of a round-robin chess tournament between five players: \(H, I, J, K\) and \(L\). In each game, there is a winner and a loser.

Two games still need to be played.

\begin{aligned}
&\quad \quad \quad\quad \quad \quad\quad \quad \quad \quad \quad \quad \   \textit{loser}\\
&\quad \quad\quad \quad \quad \ \ \  H \quad \quad  \ \ \   I \quad \quad \quad J \quad \quad \ \ \ K \quad \quad  \quad L \\
& \textit{winner} \quad \begin{array}{ccccc}
H\\
\\
I\\
\\
J\\
\\
K\\
\\
L
\end{array}
\begin {bmatrix}
0 \quad & \quad 1 \quad & \quad 0 \quad & \quad 1 \quad & \quad 0 \\
\\
0 \quad& 0 & \ldots & 1 & \quad \ldots \\
\\
1 \quad& \ldots & 0 & 1 & \quad 0 \\
\\
0 \quad& 0 & 0 & 0 & \quad 1 \\
\\
1 \quad& \ldots & 1 & 0 & \quad 0
\end{bmatrix}\\
&
\end{aligned}

A '1' in the matrix shows that the player named in that row defeated the player named in that column. For example, the 1 in row 4 shows that player \(K\) defeated player \(L\).

A '...' in the matrix shows that the player named in that row has not yet competed against the player in that column.

At the end of the tournament, players will be ranked by calculating the sum of their one-step and two-step dominances.

The player with the highest sum will be ranked first. The player with the second-highest sum will be ranked second, and so on.

Which one of the following is not a potential outcome after the final two games have been played?

  1. Player \(I\) will be ranked first.
  2. Player \(I\) will be ranked fifth.
  3. Player \(J\) will be ranked first.
  4. Player \(J\) will be ranked fifth.
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Games to be played:}\ \ I\ \text{vs}\ J, \ \ I\ \text{vs}\ L\)

\(\text{Let the results matrix be}\ D\)

♦ Mean mark 53%.

\(\text{Case 1: Player}\ I\ \text{wins both games}\)

\(D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 1 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0 
\end{bmatrix}+
 \begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 1 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0 
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 1 & 2 & 2 \\
\ 2 & 0 & 2 & 2 & 2 \\
\ 1 & 1 & 0 & 2 & 1 \\
\ 1 & 0 & 1 & 0 & 1 \\
\ 2 & 1 & 1 & 2 & 0 
\end{bmatrix}\begin{array}{ccccc}
\ \ H=6\\
\ \ I=8\\
\ \ J=5\\
\ \ K=3\\
\ \ L=6
\end{array}\)
  

\(\text{Case 2: Player}\ I\ \text{loses to}\ J, \text{defeats}\ L\)

\(D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 1 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0 
\end{bmatrix}+
 \begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 1 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 0 & 1 & 0 & 0 
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 0 & 2 & 2 \\
\ 1 & 0 & 1 & 1 & 2 \\
\ 1 & 2 & 0 & 3 & 2 \\
\ 1 & 0 & 1 & 0 & 1 \\
\ 2 & 2 & 1 & 2 & 0 
\end{bmatrix}\begin{array}{ccccc}
\ \ H=5\\
\ \ I=5\\
\ \ J=8\\
\ \ K=3\\
\ \ L=7
\end{array}\)
  

\(\text{Case 3: Player}\ I\ \text{defeats}\ J, \text{loses to}\ L\)

\(D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 0 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0 
\end{bmatrix}+
 \begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 1 & 1 & 0 \\
\ 1 & 0 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 1 & 2 & 1 \\
\ 1 & 0 & 1 & 2 & 1 \\
\ 1 & 1 & 0 & 2 & 1 \\
\ 1 & 1 & 1 & 0 & 1 \\
\ 2 & 2 & 2 & 3 & 0 
\end{bmatrix}\begin{array}{ccccc}
\ \ H=5\\
\ \ I=5\\
\ \ J=5\\
\ \ K=4\\
\ \ L=9
\end{array}\)
  

\(\text{Case 4: Player}\ I\ \text{loses both games}\)

\(D+D^{2} =\begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 0 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0 
\end{bmatrix}+
 \begin{bmatrix}
\ 0 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 1 & 0 \\
\ 1 & 1 & 0 & 1 & 0 \\
\ 0 & 0 & 0 & 0 & 1 \\
\ 1 & 1 & 1 & 0 & 0
\end{bmatrix}^{2}
=\begin{bmatrix}
\ 0 & 1 & 0 & 2 & 1 \\
\ 0 & 0 & 0 & 1 & 1 \\
\ 1 & 2 & 0 & 3 & 1 \\
\ 1 & 1 & 1 & 0 & 1 \\
\ 2 & 3 & 1 & 3 & 0 
\end{bmatrix}\begin{array}{ccccc}
\ \ H=4\\
\ \ I=2\\
\ \ J=7\\
\ \ K=4\\
\ \ L=9
\end{array}\)
  

\(\text{In all cases Player }I\ \text{or}\ K\ \text{will be ranked 5th.}\)

\(\therefore\ \text{Option D is not a potential outcome.}\)

\(\Rightarrow D\)

Matrices, GEN1 2024 VCAA 29 MC

A tennis team consists of five players: Quinn, Rosie, Siobhan, Trinh and Ursula.

When the team competes, players compete in the order of first, then second, then third, then fourth.

The fifth player has a bye (does not compete).

On week 1 of the competition, the players competed in the following order.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \quad\textbf{First} \quad\rule[-1ex]{0pt}{0pt}& \quad \textbf{Second} \quad& \quad\textbf{Third} \quad& \quad\textbf{Fourth}\quad & \quad\textbf{Bye} \quad\\
\hline
\rule{0pt}{2.5ex} \text{Quinn} \rule[-1ex]{0pt}{0pt}& \text {Rosie} & \text {Siobhan} & \text { Trinh } & \text {Ursula} \\
\hline
\end{array}

This information can be represented by matrix \(G_1\), shown below.

\(G_1=\begin{bmatrix} Q & R & S & T & U \end{bmatrix}\)

Let \(G_n\) be the order of play in week \(n\).

The playing order changes each week and can be determined by the rule  \(G_{n+1}=G_n \times P\)

\(\text{where}\quad \\P=\begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0  \\
0 & 0 & 0 & 1 & 0 \end{bmatrix}\)

Which player has a bye in week 4 ?

  1. Quinn
  2. Rosie
  3. Siobhan
  4. Trinh
Show Answers Only

\(C\)

Show Worked Solution

\(G_1=\begin{bmatrix} Q & R & S & T & U \end{bmatrix}\)

\(G_2=\begin{bmatrix} S & Q & T & U & R \end{bmatrix}\)

\(G_3=\begin{bmatrix} T & S & U & R & Q \end{bmatrix}\)

\(G_4=\begin{bmatrix} U & T & R & Q & S \end{bmatrix}\)

\(\text{5th player has a bye.}\)

\(\therefore\ \text{Siobhan does not compete in week 4}\)

\(\Rightarrow C\)

♦ Mean mark 49%.

Matrices, GEN1 2024 VCAA 27 MC

Consider the following matrix, where  \(h \neq 0\).

\begin{bmatrix}
4 & g \\
8 & h
\end{bmatrix}

The inverse of this matrix does not exist when \(g\) is equal to

  1. \(-2 h\)
  2. \(\dfrac{h}{2}\)
  3. \(h\)
  4. \(2 h\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Inverse will not exist if determinant = 0}\)

\(\text{Find}\ g\ \text{when}\ \ 4h-8g=0:\)

\(g=\dfrac{h}{2}\)

\(\Rightarrow B\)

♦ Mean mark 50%.

Recursion and Finance, GEN1 2024 VCAA 24 MC

André invested $18 000 in an account for five years, with interest compounding monthly.

He adds an extra payment into the account each month immediately after the interest is calculated.

For the first two years, the balance of the account, in dollars, after \(n\) months, \(A_n\), can be modelled by the recurrence relation

\(A_0=18\,000, \quad A_{n+1}=1.002 A_n+100\)

After two years, André decides he would like the account to reach a balance of $30 000 at the end of the five years.

He must increase the value of the monthly extra payment to achieve this.

The minimum value of the new payment for the last three years is closest to

  1. $189.55
  2. $195.45
  3. $202.35
  4. $246.55
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Step 1: Using CAS}\)

\(\text{Annual interest rate}\ = (1.002-1) \times 12 =  0.024 = 2.4\% \)

\(\text{Compounding periods (2 years)}\ = 2 \times 12=24\)

\(\text{Value of investment after 2 years}\ =$21\,340.18\)

♦ Mean mark 46%.

\(\text{Step 2: Using CAS}\)

\(\text{Next 3 years:}\ N=3 \times 12 = 36\)

\(\text{Minimum value of new payment}\ =$189.55\)

\(\Rightarrow A\)

Recursion and Finance, GEN1 2024 VCAA 22-23 MC

Stewart takes out a reducing balance loan of \$240 000, with interest calculated monthly.

Stewart makes regular monthly repayments.

Three lines of the amortisation table are shown below.

\begin{array}{|c|r|r|r|r|}
\hline
\rule{0pt}{2.5ex} \textbf{Payment} & \textbf {Payment} & \textbf {Interest} &\textbf{Principal reduction} & \textbf{Balance}\\
\rule[-1ex]{0pt}{0pt}\textbf{number} & \textbf{(\($\))}\ \ \ \ \  & \textbf{(\($\))}\ \ \ \ \ & \textbf{(\($\))}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ & \textbf{(\($\))}\ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} 0 & 0.00 & 0.00 & 0.00\ \ \ \ \ \ \ \ \ \  & 240\,000.00 \\
\hline
\rule{0pt}{2.5ex} 1 & 2741.05 & 960.00 & 1781.05\ \ \ \ \ \ \ \ \ \  & 238\,218.95 \\
\hline
\rule{0pt}{2.5ex} 2 & 2741.05 & & & \\
\hline
\end{array}

 
Part 1

The principal reduction associated with Payment number 2 is closest to

  1. $1773.93
  2. $1781.05
  3. $1788.17
  4. $2741.05

 
Part 2

The number of years that it will take Stewart to repay the loan in full is closest to

  1. 9
  2. 10
  3. 11
  4. 12
Show Answers Only

Part 1: \(C\)

Part 2: \(A\)

Show Worked Solution

Part 1

\(\text{Interest rate (monthly)}\ =\dfrac{960}{240\,000} = 0.004\)

\(\text{Interest repaid (payment 2)}\ =\dfrac{960}{240\,000}\times 238\,218.95=$952.88\)

\(\text{Principal Reduction (payment 2)}\ =2741.05-952.88=$1788.17\)

\(\Rightarrow C\)
 

Part 2

\(\text{Using CAS:}\)

\(\text{Annual interest rate}\ = 12 \times 0.004 = 4.8\%\)

\(\text{Number of repayments }= 108\ \text{months} = 9\ \text{years}\)

\(\Rightarrow A\)

♦ Mean mark (Part 2) 54%.

Recursion and Finance, GEN1 2024 VCAA 20 MC

Dainika invested $2000 for three years at 4.4% per annum, compounding quarterly.

To earn the same amount of interest in three years in a simple interest account, the annual simple interest rate would need to be closest to

  1. 4.60%
  2. 4.68%
  3. 4.84%
  4. 4.98%
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Interest rate (per compounding period)} = \dfrac{4.4}{4} = 1.1 \%\)

\(\text{Compounding periods}\ =3\times4=12\)

\(FV=PV(1+r)^{n} = 2000(1+0.011)^{12}=$2280.57\)

\(\text{Annual interest}\ = \dfrac{280.57}{3}=$93.52\)

\(\text{S.I. rate}\ =\dfrac{93.52}{2000}\times 100\%=4.676\dots\%\)

\(\Rightarrow B\)

♦ Mean mark 48%.

Recursion and Finance, GEN1 2024 VCAA 17 MC

A first-order linear recurrence relation of the form

\(u_0=a, \quad \quad u_{n+1}=Ru_n+d\)

generates the terms of a sequence. A geometric sequence will be generated if

  1. \(R=1\)  and  \(d=-1\)
  2. \( R=1\)  and  \(d=1\)
  3. \(R=4\)  and  \(d=-1\)
  4. \(R=2\)  and  \(d=0\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Consider Option D where }R=2\ \text{and }d=0\)

\(u_0=a, \quad u_{n+1}=2u_n\)

\(\text{Generates the geometric sequence }\ a,\ 2a,\ 4a,\ 8a,\ …\)

  
\(\Rightarrow D\)

♦♦ Mean mark 35%.

Data Analysis, GEN1 2024 VCAA 11-12 MC

The number of breeding pairs of a small parrot species has been declining over recent years.

The table below shows the number of breeding pairs counted, pairs, and the year number, year, for the last 12 years. A scatterplot of this data is also provided.

The association between pairs and year is non-linear.
 

Part 1

The scatterplot can be linearised using a logarithmic (base 10) transformation applied to the explanatory variable.

The least squares equation calculated from the transformed data is closest to

  1. \(\log _{10}(pairs)=2.44-0.0257 \times year\)
  2. \(\log _{10}(pairs)=151-303 \times year\)
  3. \(pairs =274-12.3 \times \log _{10}(year)\)
  4. \(pairs =303-151 \times \log _{10}( year)\)

 
Part 2

A reciprocal transformation applied to the variable \(pairs\) can also be used to linearise the scatterplot.

When a least squares line is fitted to the plot of  \(\dfrac{1}{pairs}\)  versus \(year\), the largest difference between the actual value and the predicted value occurs at \(year\)

  1. 1
  2. 2
  3. 11
  4. 12
Show Answers Only

Part 1: \(D\)

Part 2: \(A\)

Show Worked Solution

Part 1

\(\text{Apply reciprocal transformation and find regression line:}\)

\(\text{Least squares equation:}\ \ pairs=303-151\times\log_{10}{(year)}\)

\(\Rightarrow D\)
 

Part 2

\(\text{Largest difference occurs at year 1}\)

\(\Rightarrow A\)

♦ Mean mark (Part 1) 48%.

Data Analysis, GEN1 2024 VCAA 9-10 MC

The least squares equation for the relationship between the average number of male athletes per competing nation, males, and the number of the Summer Olympic Games, number, is

\(males =67.5-1.27 \times number\)
 

Part 1

The summary statistics for the variables number and males are shown in the table below.
 

The value of Pearson's correlation coefficient, \(r\), rounded to three decimal places, is closest to

  1. \(-0.569\)
  2. \(-0.394\)
  3. \(0.394\)
  4. \(0.569\)

 
Part 2

At which Summer Olympic Games will the predicted average number of males be closest to 25.6 ?

  1. 31st
  2. 32nd
  3. 33rd
  4. 34th
Show Answers Only

Part 1: \(A\)

Part 2: \(C\)

Show Worked Solution

Part 1

\(b\) \(=r \times \dfrac{s_y}{s_x}\)
\(-1.27\) \(=r\times\dfrac{19}{8.51}\)
\(\therefore\ r\) \(=\dfrac{-1.27\times 8.51}{19}\)
  \(=-0.5688\dots\)

 
\(\Rightarrow A\)
 

♦ Mean mark (Part 1) 52%.

Part 2

\(males\) \(=67.5-1.27\times number\)
\(25.6\) \(=67.5-1.27\times number\)
\(number\) \(=\dfrac{67.5-25.6}{1.27}\)
  \(=32.992\dots\)
  \(\approx 33\)

 

\(\Rightarrow C\)

Data Analysis, GEN1 2024 VCAA 6 MC

More than 11 000 athletes from more than 200 countries competed in the Tokyo Summer Olympic Games.

An analysis of the number of athletes per country produced the following five-number summary.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Minimum} \rule[-1ex]{0pt}{0pt}& \textbf{First quartile } & \textbf{Median } & \textbf{Third quartile} & \textbf{Maximum } \\
\hline
\rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt}& 5 & 11 & 48 & 613 \\
\hline
\end{array}

The smallest number of athletes per country that would display as an outlier on a boxplot of this data is

  1. 49
  2. 112
  3. 113
  4. 613
Show Answers Only

\(C\)

Show Worked Solution

\(IQR=48-5=43\)

\(\text{Upper boundary}\) \(=Q_3+1.5\times IQR\)
  \(=48+1.5\times 43\)
  \(=112.5\)

 
\(\therefore\ \text{Smallest number of athletes to show as an outlier = 113.}\)

\(\Rightarrow C\)

♦ Mean mark 53%.

Complex Numbers, EXT2 N2 2024 HSC 16b

The number  \(w=e^{\small{\dfrac{2 \pi i}{3}}}\)  is a complex cube root of unity. The number \(\gamma\) is a cube root of \(w\).

  1. Show that  \(\gamma+\bar{\gamma}\)  is a real root of  \(z^3-3 z+1=0\).   (3 marks)

  2. By using part (i) to find the exact value of  \(\cos \dfrac{2 \pi}{9} \cos \dfrac{4 \pi}{9} \cos \dfrac{8 \pi}{9}\), deduce the value(s) of  \(\cos \dfrac{2^n \pi}{9} \cos \dfrac{2^{n+1} \pi}{9} \cos \dfrac{2^{n+2} \pi}{9}\)  for all integers  \(n \geq 1\). Justify your answer.   (3 marks)

Show Answers Only

i.     \(w=e^{\small{\dfrac{2 \pi i}{3}}} \Rightarrow \ \text{complex (cube) root of 1}\)

\(\gamma^3=w, \quad \gamma \bar{\gamma}=\abs{\gamma}^2=1\)

\(\text{Show}\ \ \gamma+\bar{\gamma}\ \ \text{is a real root of} \ \ z^3-3 z+1=0:\)

  \((\gamma+\bar{\gamma})^3-3(\gamma+\bar{\gamma})+1\)
    \(=\gamma^3+3 \gamma^2 \bar{\gamma}+3 \gamma \bar{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=\gamma^3+3 \gamma\abs{\gamma}^2+3 \bar{\gamma}\abs{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+3 \gamma+3 \bar{\gamma}+\bar{w}-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+\bar{w}+1\)
    \(=0 \quad \text{(Sum of complex roots of 1 = 0)}\)

 
\(\text{Show}\ \ \gamma+\bar{\gamma} \ \ \text{is real:}\)

\(\gamma+\bar{\gamma}=a+b i+a-b i=2 a \quad(a \in \mathbb{R})\)

\(\therefore \gamma+\bar{\gamma} \ \ \text{is a real root of} \ \  z^3-3 z+1=0.\)
 

ii.   \(e^{\small{\dfrac{2 \pi i}{9}}}\ \ \text{is a cube root of}\ \ \omega \ \Rightarrow \ \Bigg(e^{\small{\dfrac{2 \pi i}{9}}}\Bigg)^3=e^{\small{\dfrac{2 \pi i}{3}}}\)

\(\text{Cubic roots are} \ \dfrac{2}{3} \ \text {rotations from each other.}\)

\(\text{Roots of} \ \ z^3-3 z+1=0:\)

  \((\gamma+\bar{\gamma})_1\) \(=2 \cos \left(\dfrac{2 \pi}{9}\right)\)
  \((\gamma+\bar{\gamma})_2\) \(=2 \cos \left(\dfrac{8 \pi}{9}\right)\)
  \((\gamma+\bar{\gamma})_3\) \(=2 \cos \left(\dfrac{4\pi}{9}\right)\)

 

\(\text{Product of roots} \ \ \alpha B \gamma=-\dfrac{d}{a}:\)

  \(2 \cos \left(\dfrac{2 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{8 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{4 \pi}{4}\right)\) \(=-1\)
  \(\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right)\) \(=-\dfrac{1}{8}\)

 
\(\text{Consider} \ \ \cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right):\)

\(\text{If} \ \ n=1,\)

\(\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right)=-\dfrac{1}{8}\ \ \text{(see above)}\)

\(\text {If} \ \ n=2,\)

\(\cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{2\pi}{9}\right)=-\dfrac{1}{8}\)

 
\(\Rightarrow \ \text{Multiplying each argument by 2 does not change the equation}\)

\(\therefore \cos \left(\dfrac{2^n \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right)=-\dfrac{1}{8}\)

Show Worked Solution

i.     \(w=e^{\small{\dfrac{2 \pi i}{3}}} \Rightarrow \ \text{complex (cube) root of 1}\)

\(\gamma^3=w, \quad \gamma \bar{\gamma}=\abs{\gamma}^2=1\)

\(\text{Show}\ \ \gamma+\bar{\gamma}\ \ \text{is a real root of} \ \ z^3-3 z+1=0:\)

♦♦ Mean mark (i) 36%.
  \((\gamma+\bar{\gamma})^3-3(\gamma+\bar{\gamma})+1\)
    \(=\gamma^3+3 \gamma^2 \bar{\gamma}+3 \gamma \bar{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=\gamma^3+3 \gamma\abs{\gamma}^2+3 \bar{\gamma}\abs{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+3 \gamma+3 \bar{\gamma}+\bar{w}-3 \gamma-3 \bar{\gamma}+1\)
    \(=w+\bar{w}+1\)
    \(=0 \quad \text{(Sum of complex roots of 1 = 0)}\)

 

\(\text{Show}\ \ \gamma+\bar{\gamma} \ \ \text{is real:}\)

\(\gamma+\bar{\gamma}=a+b i+a-b i=2 a \quad(a \in \mathbb{R})\)

\(\therefore \gamma+\bar{\gamma} \ \ \text{is a real root of} \ \  z^3-3 z+1=0.\)
 

ii.   \(e^{\small{\dfrac{2 \pi i}{9}}}\ \ \text{is a cube root of}\ \ \omega \ \Rightarrow \ \Bigg(e^{\small{\dfrac{2 \pi i}{9}}}\Bigg)^3=e^{\small{\dfrac{2 \pi i}{3}}}\)

\(\text{Cubic roots are} \ \dfrac{2}{3} \ \text {rotations from each other.}\)

♦♦♦ Mean mark (ii) 18%.

\(\text{Roots of} \ \ z^3-3 z+1=0:\)

  \((\gamma+\bar{\gamma})_1\) \(=2 \cos \left(\dfrac{2 \pi}{9}\right)\)
  \((\gamma+\bar{\gamma})_2\) \(=2 \cos \left(\dfrac{8 \pi}{9}\right)\)
  \((\gamma+\bar{\gamma})_3\) \(=2 \cos \left(\dfrac{4\pi}{9}\right)\)

 

\(\text{Product of roots} \ \ \alpha B \gamma=-\dfrac{d}{a}:\)

  \(2 \cos \left(\dfrac{2 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{8 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{4 \pi}{4}\right)\) \(=-1\)
  \(\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right)\) \(=-\dfrac{1}{8}\)

 
\(\text{Consider} \ \ \cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right):\)

\(\text{If} \ \ n=1,\)

\(\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right)=-\dfrac{1}{8}\ \ \text{(see above)}\)

\(\text {If} \ \ n=2,\)

\(\cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{2\pi}{9}\right)=-\dfrac{1}{8}\)

 
\(\Rightarrow \ \text{Multiplying each argument by 2 does not change the equation}\)

\(\therefore \cos \left(\dfrac{2^n \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right)=-\dfrac{1}{8}\)

Calculus, EXT2 C1 2024 HSC 15d

Using a suitable substitution, find  \(\displaystyle\int \dfrac{2 x^2}{\sqrt{2 x-x^2}}\, d x\).   (3 marks)

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

Show Worked Solution

  \(\displaystyle \int \dfrac{2x^2}{\sqrt{2x-x^2}}\, dx\) \(=\displaystyle \int \frac{2 x^2}{\sqrt{1-(1-2 x+x^2)}}\, dx\)
    \(=\displaystyle \int \frac{2 x^2}{\sqrt{1-(x-1)^2}} \, dx\)
♦ Mean mark 49%.

\(\text {Let} \ \ x-1=\sin \theta \ \Rightarrow \ x=1+\sin \theta\)

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

  \(I\) \(=\displaystyle \int \frac{2(1+\sin \theta)^2}{\sqrt{1-\sin ^2 \theta}} \cdot \cos \theta \, d \theta\)
    \(=2 \displaystyle \int \frac{1+2 \sin \theta+\sin ^2 \theta}{\cos \theta} \cdot \cos \theta \, d \theta\)
    \(=2 \displaystyle \int 1+2 \sin \theta+\frac{1}{2}(1-\cos (2 \theta)) \, d \theta\)
    \(=\displaystyle \int 2+4 \sin \theta+1-\cos (2 \theta) \, d \theta\)
    \(=\displaystyle \int 3+4 \sin \theta-\cos (2 \theta) \, d \theta\)
    \(=3 \theta-4 \cos \theta-\dfrac{1}{2} \sin (2 \theta)+c\)
    \(=3 \theta-4 \cos \theta-\sin \theta \cos \theta+c\)

 

\(\Rightarrow \cos \theta=\sqrt{1^2-(x-1)^2}=\sqrt{2x-x^2}\)
 

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

Vectors, EXT2 V1 2024 HSC 14e

The diagram shows triangle \(O Q A\).

The point \(P\) lies on \(O A\) so that  \(O P: O A=3: 5\).

The point \(B\) lies on \(O Q\) so that  \(O B: O Q=1: 3\).

The point \(R\) is the intersection of \(A B\) and \(P Q\).

The point \(T\) is chosen on \(A Q\) so that \(O, R\) and \(T\) are collinear.
 

Let  \(\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}\)  and  \(\overrightarrow{P R}=k \overrightarrow{P Q}\)  where \(k\) is a real number.

  1. Show that  \(\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}\).   (2 marks)

Writing  \(\overrightarrow{A R}=h \overrightarrow{A B}\), where \(h\) is a real number, it can be shown that  \(\overrightarrow{O R}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b}\).  (Do NOT prove this.)

  1. Show that  \(k=\dfrac{1}{6}\).   (2 marks)

  2. Find \(\overrightarrow{O T}\) in terms of \(\underset{\sim}{a}\) and \(\underset{\sim}{b}\).   (2 marks)

Show Answers Only

i.    \(\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}, \ \overrightarrow{P R}=k \overrightarrow{P Q}\)

\(\overrightarrow{O P}: \overrightarrow{O A}=3: 5 \ \Rightarrow \ \overrightarrow{O P} = \dfrac{3}{5} \times \overrightarrow{O A} \)

\(\overrightarrow{O B}: \overrightarrow{O Q}=1: 3\ \Rightarrow \ \overrightarrow{O Q} = 3 \times \overrightarrow{OB} \)

\(\text {Show } \overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}\)

  \(\overrightarrow{O R}\) \(=\overrightarrow{O P}+k \overrightarrow{P Q}\)
    \(=\overrightarrow{O P}+k(\overrightarrow{O Q}-\overrightarrow{O P})\)
    \(=(1-k) \overrightarrow{O P}+k \overrightarrow{O Q}\)
    \(=\dfrac{3}{5}(1-k)\underset{\sim}{a}+3 k \underset{\sim}{b}\)

 

ii.    \(\overrightarrow{A R}=h \overrightarrow{A B}, \quad h \in \mathbb{R}\)

\(\overrightarrow{OR}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b} \ \ \text{(given)}\)

\(\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b} \ \ \text{(part(i))}\)

\(\text{Vector }\underset{\sim}{a} \neq \lambda \underset{\sim}{b}\ (\lambda \in \mathbb{R})\ \Rightarrow \ \text{linearly independent and are basis vectors for } \overrightarrow{O R}\).

\(\text {Equating coefficients:}\)

   \(\dfrac{3}{5}(1-k)=1-h\ \ldots\ (1)\)

   \(h=3 k\ \ldots\ (2)\)

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

     \(\dfrac{3}{5}(1-k)\) \(=1-3 k\)
  \(3-3 k\) \(=5-15 k\)
  \(k\) \(=\dfrac{1}{6}\)

 
iii.    \(\overrightarrow{O T}=\dfrac{3}{4}(\underset{\sim}{a}+\underset{\sim}{b})\)

Show Worked Solution

i.    \(\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}, \ \overrightarrow{P R}=k \overrightarrow{P Q}\)

\(\overrightarrow{O P}: \overrightarrow{O A}=3: 5 \ \Rightarrow \ \overrightarrow{O P} = \dfrac{3}{5} \times \overrightarrow{O A} \)

\(\overrightarrow{O B}: \overrightarrow{O Q}=1: 3\ \Rightarrow \ \overrightarrow{O Q} = 3 \times \overrightarrow{OB} \)

\(\text {Show } \overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}\)

  \(\overrightarrow{O R}\) \(=\overrightarrow{O P}+k \overrightarrow{P Q}\)
    \(=\overrightarrow{O P}+k(\overrightarrow{O Q}-\overrightarrow{O P})\)
    \(=(1-k) \overrightarrow{O P}+k \overrightarrow{O Q}\)
    \(=\dfrac{3}{5}(1-k)\underset{\sim}{a}+3 k \underset{\sim}{b}\)

  

ii.    \(\overrightarrow{A R}=h \overrightarrow{A B}, \quad h \in \mathbb{R}\)

\(\overrightarrow{OR}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b} \ \ \text{(given)}\)

\(\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b} \ \ \text{(part(i))}\)

\(\text{Vector }\underset{\sim}{a} \neq \lambda \underset{\sim}{b}\ (\lambda \in \mathbb{R})\ \Rightarrow \ \text{linearly independent and are basis vectors for } \overrightarrow{O R}\).

\(\text {Equating coefficients:}\)

   \(\dfrac{3}{5}(1-k)=1-h\ \ldots\ (1)\)

   \(h=3 k\ \ldots\ (2)\)

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

     \(\dfrac{3}{5}(1-k)\) \(=1-3 k\)
  \(3-3 k\) \(=5-15 k\)
  \(k\) \(=\dfrac{1}{6}\)

 

iii.    \(\overrightarrow{O T}=\lambda \overrightarrow{O R}\)

\(\text {Using parts (i) and (ii):}\)

\(\overrightarrow{O R}=\dfrac{3}{5}\left(1-\dfrac{1}{6}\right)\underset{\sim}{a}+3\left(\dfrac{1}{6}\right) \underset{\sim}{b}=\dfrac{1}{2}\left(\underset{\sim}{a}+\underset{\sim}{b}\right)\)

\(\overrightarrow{O T}=\dfrac{\lambda}{2}(\underset{\sim}{a}+\underset{\sim}{b})\)

♦ Mean mark (iii) 45%.
 

\(\text {Find } \lambda:\)

  \(\overrightarrow{O T}\) \(=\overrightarrow{O A}+\mu \overrightarrow{A Q}\)
  \(\dfrac{\lambda}{2}(\underset{\sim}{a}+\underset{\sim}{b})\) \(=\underset{\sim}{a}+\mu(3 \underset{\sim}{b}-\underset{\sim}{a}) \quad \Big(\text{noting}\ \overrightarrow{A Q}=\overrightarrow{O Q}-\overrightarrow{O A}=3 \underset{\sim}{b}-\underset{\sim}{a}\Big)\)
    \(=\underset{\sim}{a}(1-\mu)+3 \mu \underset{\sim}{b}\)

 
\(\text {Equating coefficients:}\)

\(\dfrac{\lambda}{2}=3 \mu \ \Rightarrow \ \mu=\dfrac{\lambda}{6}\)

  \(1-\dfrac{\lambda}{6}\) \(=\dfrac{\lambda}{2}\)
  \(6-\lambda\) \(=3 \lambda\)
  \(\lambda\) \(=\dfrac{3}{2}\)

 
\(\therefore \overrightarrow{O T}=\dfrac{3}{4}(\underset{\sim}{a}+\underset{\sim}{b})\)

Complex Numbers, EXT2 N2 2024 HSC 14c

For the complex numbers \(z\) and \(w\), it is known that  \(\arg \left(\dfrac{z}{w}\right)=-\dfrac{\pi}{2}\).  

Find \(\left|\dfrac{z-w}{z+w}\right|\).   (2 marks)

Show Answers Only

\(\abs{\dfrac{z-w}{z+w}}=1\)

Show Worked Solution

\(\arg \left(\dfrac{z}{w}\right)=\arg \, z-\arg \, w=-\dfrac{\pi}{2}\)

\(\text{Graphically, \(\arg \, z\) is a \(90^{\circ}\) anticlockwise rotation from \(\arg \, w\).}\)

♦ Mean mark 49%.

\(\abs{z-w}=\abs{z+w} \ \text{(diagonals of rectangle)}\)

\(\therefore \abs{\dfrac{z-w}{z+w}}=1\)

Proof, EXT2 P1 2024 HSC 14d

The following argument attempts to prove that  \(0=1\).

  1. We evaluate \(\displaystyle\int \frac{1}{x}\, d x\) using the method of integration by parts.
  \(\displaystyle \int \frac{1}{x}\,d x\) \(=\displaystyle \int \frac{1}{x} \times 1\, d x\)
    \(=\displaystyle\frac{1}{x} \times x-\int-\frac{1}{x^2} x\, d x\)
    \(=1+\displaystyle\int \frac{1}{x}\, d x\)
  1.  
    We may now subtract \(\displaystyle \int \frac{1}{x}\,d x\) from both sides to show that  \(0=1\).

Explain what is wrong with this argument.  (2 marks)

Show Answers Only

\(\text {Consider the line in proof }\)

\(\displaystyle {\int \frac{1}{x}\, d x=1+\int \frac{1}{x}\, d x}\)

 \(\Rightarrow \text{ Each integral will have its own constant }\)

\(\Rightarrow \text{If constants are \(c_1\) and \(c_2, \abs{c_1-c_2}=1\) in this proof.}\)

\(\Rightarrow \text{Subtracting}\ \displaystyle \int \frac{1}{x}\,d x\ \text{from both sides invalidates the proof.}\)

Show Worked Solution

\(\text {Consider the line in proof }\)

\(\displaystyle {\int \frac{1}{x}\, d x=1+\int \frac{1}{x}\, d x}\)

♦♦ Mean mark 34%.

\(\Rightarrow \text{ Each integral will have its own constant }\)

\(\Rightarrow \text{If constants are \(c_1\) and \(c_2, \abs{c_1-c_2}=1\) in this proof.}\)

\(\Rightarrow \text{Subtracting}\ \displaystyle \int \frac{1}{x}\,d x\ \text{from both sides invalidates the proof.}\)

Vectors, EXT2 V1 2024 HSC 13a

The point \(A\) has position vector  \(8 \underset{\sim}{i}-6 \underset{\sim}{j}+5 \underset{\sim}{k}\). The line \(\ell\) has vector equation

\(x \underset{\sim}{i}+y \underset{\sim}{j}+z \underset{\sim}{k}=t(\underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k})\).

The point \(B\) lies on \(\ell\) and has position vector  \(p \underset{\sim}{i}+p \underset{\sim}{j}+2 p \underset{\sim}{k}\).

  1. Show that  \(\abs{A B}^2=6 p^2-24 p+125\).   (1 mark)

  2. Hence, or otherwise, determine the shortest distance between the point \(A\) and the line \(\ell\).  (2 marks)

Show Answers Only

i.     \(A\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right), \quad \ell: \left(\begin{array}{l}x \\ y \\ z\end{array}\right)+t\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right), \quad B\left(\begin{array}{c}p \\ p \\ 2 p\end{array}\right)\)
 

\(\overrightarrow{A B}=\left(\begin{array}{l}p \\ p \\ 2 p\end{array}\right)-\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right)=\left(\begin{array}{c}p-8 \\ p+6 \\ 2 p-5\end{array}\right)\)
 

  \(\abs{AB}^2\) \(=(p-8)^2+(p+6)^2+(2 p-5)^2\)
    \(=p^2-16 p+64+p^2+12 p+36+4 p^2-20 p+25\)
    \(=6 p^2-24 p+125\)

 

ii.    \(\abs{AB}_{\text {min}}=\sqrt{101} \text { units}\)

Show Worked Solution

i.     \(A\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right), \quad \ell: \left(\begin{array}{l}x \\ y \\ z\end{array}\right)+t\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right), \quad B\left(\begin{array}{c}p \\ p \\ 2 p\end{array}\right)\)
 

\(\overrightarrow{A B}=\left(\begin{array}{l}p \\ p \\ 2 p\end{array}\right)-\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right)=\left(\begin{array}{c}p-8 \\ p+6 \\ 2 p-5\end{array}\right)\)
 

  \(\abs{AB}^2\) \(=(p-8)^2+(p+6)^2+(2 p-5)^2\)
    \(=p^2-16 p+64+p^2+12 p+36+4 p^2-20 p+25\)
    \(=6 p^2-24 p+125\)

  

ii.    \(\text{Find shortest distance between \(A\) and \(\ell\).}\)

\(\Rightarrow \text { Find \(p\) when \(\abs{A B}\) is a minimum:}\)

\(\text{Minimum occurs when}\ \ p=\dfrac{-b}{2 a}=\dfrac{24}{2 \times 6}=2\)

\(\therefore \abs{AB}_{\text {min}}=\sqrt{6(2)^2-24(2)+125}=\sqrt{101} \text { units}\)

♦ Mean mark (ii) 45%.

Proof, EXT2 P1 2024 HSC 12d

Explain why there is no integer \(n\) such that  \((n+1)^{41}-79 n^{40}=2\).   (2 marks)

Show Answers Only

\(\text{Show}\ \ \nexists \, n  \in \mathbb{Z}: \ (n+1)^{41}-79 n^{40}=2\)

\(\text {If \(n\) is even:}\)

\(\text{LHS}=(\text{odd})^{41}-79(\text{even})^{40}=\text {odd}-\text {even}=\text {odd} \neq 2\)

\(\text {If \(n\) is odd:}\)

\(\text {LHS }=(\text {even})^{41}-79(\text {odd})^{40}=\text {even}- \text {odd}=\text {odd} \neq 2\)

\(\therefore\ \text {By contradiction}\)

\(\nexists \, n \in \mathbb{Z}: \ (n+1)^{41}-79 n^{40}=2\)

Show Worked Solution

\(\text{Show}\ \ \nexists \, n  \in \mathbb{Z}: \ (n+1)^{41}-79 n^{40}=2\)

\(\text {If \(n\) is even:}\)

\(\text{LHS}=(\text{odd})^{41}-79(\text{even})^{40}=\text {odd}-\text {even}=\text {odd} \neq 2\)

\(\text {If \(n\) is odd:}\)

\(\text {LHS }=(\text {even})^{41}-79(\text {odd})^{40}=\text {even}- \text {odd}=\text {odd} \neq 2\)

\(\therefore\ \text {By contradiction}\)

\(\nexists \, n \in \mathbb{Z}: \ (n+1)^{41}-79 n^{40}=2\)

♦ Mean mark 49%.

Trigonometry, EXT1 T3 2024 HSC 14c

  1. Explain why the equation  \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta\), where  \(-\pi<\theta<\pi\), has exactly one solution.   (1 marks)

  2. Solve  \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\dfrac{3 \pi}{4}\).   (2 marks)

Show Answers Only

i.     \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta \quad-\pi<\theta<\pi\)

\(\text {Range:}\ \ \tan ^{-1}(3 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), \ \tan ^{-1}(10 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\)

 \(\Rightarrow \text { Both are monotonically increasing functions}\)

\(\Rightarrow\tan ^{-1}(3 x)+\tan ^{-1}(10 x) \text{ is also monotonically increasing with range }(-\pi, \pi)\) 

 \(\Rightarrow \text{ Only 1 solution exists (horizontal line will only cut graph once).}\)
 

ii.   \(x=\dfrac{1}{2}\)

Show Worked Solution

i.     \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta \quad-\pi<\theta<\pi\)

\(\text {Range:}\ \ \tan ^{-1}(3 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), \ \tan ^{-1}(10 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\)

♦♦♦ Mean mark (i) 11%.

\(\Rightarrow \text { Both are monotonically increasing functions}\)

\(\Rightarrow\tan ^{-1}(3 x)+\tan ^{-1}(10 x) \text{ is also monotonically increasing with range }(-\pi, \pi)\) 

 \(\Rightarrow \text{ Only 1 solution exists (horizontal line will only cut graph once).}\)
 

ii.    \(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\dfrac{3 \pi}{4}\)

\(\tan \left(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)\right)=\tan \left(\dfrac{3 \pi}{4}\right)\)

♦♦ Mean mark (ii) 32%.

\(\dfrac{\tan \left(\tan ^{-1}(3 x)\right)+\tan \left(\tan ^{-1}(10 x)\right)}{1-\tan \left(\tan ^{-1}(3 x)\right) \cdot \tan \left(\tan ^{-1}(10 x)\right)}=-1\)

  \(\dfrac{3 x+10 x}{1-30 x^2}\) \(=-1\)
  \(13 x\) \(=30 x^2-1\)
  \(30 x^2-13 x-1\) \(=0\)
  \((15 x+1)(2 x-1)\) \(=0\)

 
\(x=\dfrac{1}{2}\ \ \text {or}\ \ -\dfrac{1}{15}\)

\(\text {Graph is monotonically increasing through } (0,0) \Rightarrow \ \Big(x \neq -\dfrac{1}{15} \Big)\)

\(\therefore x=\dfrac{1}{2}\)

Calculus, EXT1 C2 2024 HSC 14b

For what values of the constant \(k\) would the function  \(f(x)=\dfrac{k x}{1+x^2}+\arctan x\)  have an inverse?   (3 marks)

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\(f(x)\ \text{has an inverse for}\ \ -1 \leqslant k \leqslant 1\)

Show Worked Solution

  \(f(x)\) \(=\dfrac{k x}{1+x^2}+\arctan x\)
  \(f^{\prime}(x)\) \(=\dfrac{k\left(1+x^2\right)-k x(2 x)}{\left(1+x^2\right)^2}+\dfrac{1}{1+x^2}\)
    \(=\dfrac{k+k x^2-2 k x^2+1+x^2}{\left(1+x^2\right)^2}\)
    \(=\dfrac{x^2(1-k)+k+1}{\left(1+x^2\right)^2}\)
♦♦♦ Mean mark 26%.

\(\text{Inverse function } \Rightarrow f(x) \text { has no SPs}\)

\(x^2(1-k)+k+1 \neq 0\)

\(\text {No solution if } \ \Delta<0:\)

  \(-4(1-k)(k+1)\) \(<0\)
  \((1-k)(k+1)\) \(>0\)

 
\(f(x)\ \text{has an inverse for}\ \ -1 \leq k \leq 1\)

Calculus, EXT1 C3 2024 HSC 14a

Find the domain and range of the function that is the solution to the differential equation

\(\dfrac{d y}{d x}=e^{x+y}\)

and whose graph passes through the origin.   (4 marks)

Show Answers Only

\(\text{Domain:}\ \ x<\ln 2\)

\(\text{Range:}\ \ y>-\ln 2\)

Show Worked Solution

\(\dfrac{d y}{d x}=e^{x+y}=e^{x}\cdot e^{y}\)

  \(\displaystyle\int e^{-y}\, d y\) \(=\displaystyle \int e^x\, d x\)
  \(-e^{-y}\) \(=e^x+c\)
♦ Mean mark 52%.

\(\text{Passes through }(0,0):\)

\(-e^0=e^0+c \ \Rightarrow \ c=-2\)

  \(-e^{-y}\) \(=e^x-2\)
  \(e^{-y}\) \(=2-e^x\)
  \(-y\) \(=\ln \left(2-e^x\right)\)
  \(y\) \(=-\ln \left(2-e^x\right)\)

\(\text{Since}\ \ 2-e^x>0 \ \Rightarrow \ e^x<2\)

\(\Rightarrow \ \text{Domain:}\ \ x<\ln 2\)

\(\text{Since}\ \ e^x>0 \ \Rightarrow \ 2-e^x<2\)

\(\Rightarrow \ \text{Range:}\ \ y>-\ln 2\)

Vectors, EXT1 V1 2024 HSC 13c

The vector \(\underset{\sim}{a}\) is \(\displaystyle \binom{1}{3}\) and the vector \(\underset{\sim}{b}\) is \(\displaystyle\binom{2}{-1}\).

The projection of a vector \(\underset{\sim}{x}\) onto the vector \(\underset{\sim}{a}\) is \(k \underset{\sim}{a}\), where \(k\) is a real number.

The projection of the vector \(\underset{\sim}{x}\) onto the vector \(\underset{\sim}{b}\) is \(p \underset{\sim}{b}\), where \(p\) is a real number.

Find the vector \(\underset{\sim}{x}\) in terms of \(k\) and \(p\).   (4 marks)

Show Answers Only

\(\underset{\sim}{x}=\dfrac{5}{7} \displaystyle \binom{2 k+3 p}{4 k-p}\)

Show Worked Solution

\(\underset{\sim}{a}=\displaystyle \binom{1}{3}, \ \abs{\underset{\sim}{a}}=\sqrt{1^2+3^2}=\sqrt{10}\)

\(\underset{\sim}{b}=\displaystyle \binom{2}{-1}, \ \abs{\underset{\sim}{b}}=\sqrt{2^2+(-1)^2}=\sqrt{5}\)

\(\text{Let } \underset{\sim}{x}=\displaystyle \binom{x_1}{x_2}\)

\(\operatorname{proj}_{\underset{\sim}{a}} \underset{\sim}{x}=\dfrac{\underset{\sim}{x} \cdot \underset{\sim}{a}}{|\underset{\sim}{a}|^2} \underset{\sim}{a}=\dfrac{x_1+3 x_2}{10} \cdot \underset{\sim}{a}\)

\(k=\dfrac{x_1+3 x_2}{10} \ \Rightarrow \ x_1+3 x_2=10 k\ \ldots\ (1)\)

♦ Mean mark 47%.

\(\operatorname{proj}_{\underset{\sim}{b}} \underset{\sim}{x}=\dfrac{\underset{\sim}{b} \cdot \underset{\sim}{x}}{|\underset{\sim}{b}|^2} b=\dfrac{2 x_1-x_2}{5} \cdot \underset{\sim}{b}\)

\(p=\dfrac{2 x_1-x_2}{5} \ \Rightarrow \ 2 x_1-x_2=5 p\ \ldots\\ (2)\)
 

  \(\text {Multiply } (2) \times 3\)

\(6 x_1-3 x_2=15 p\ \ldots\ (3)\)

  \((1)+(3)\)

\(7 x_1\) \(=10 k+15 p\)  
\(x_1\) \(=\dfrac{1}{7}(10 k+15)\)  

 
\(\text {Multiply } (1) \times 2\)

\(2 x_1+6 x_2=20 k\ \ldots\ (4)\)

  \(\text {Subtract} (4)-(2)\)

\(7x_2\) \(=20 k-5 p\)  
\(x_2\) \(=\dfrac{1}{7}(20 k-5 p)\)  

 
\(\therefore \underset{\sim}{x}=\displaystyle \frac{1}{7}\binom{10 k+15 p}{20 k-5 p}=\frac{5}{7}\binom{2 k+3 p}{4 k-p}\)

Calculus, EXT1 C3 2024 HSC 13a

In an experiment, the population of insects, \(P(t)\), was modelled by the logistic differential equation

\(\dfrac{d P}{d t}=P(2000-P)\)

where \(t\) is the time in days after the beginning of the experiment.

The diagram shows a direction field for this differential equation, with the point \(S\) representing the initial population.
 

  1. Explain why the graph of the solution that passes through the point \(S\) cannot also pass through the point \(T\).   (1 mark)

  2. Clearly sketch the graph of the solution that passes through the point \(S\).   (1 mark)

  3. Find the predicted value of the population, \(P(t)\), at which the rate of growth of the population is largest.   (2 marks)

Show Answers Only

 i.    \(\text{Any solution that passes through}\ S\ \text{will follow the}\)

\(\text{slope field (not crossing any lines) and approach a horizontal}\)

\(\text{asymptote at}\ P=2000\ \text{from the lower side}\ (P<2000).\)

ii.    
       

iii.  \(P= 1000\)

Show Worked Solution

 i.    \(\text{Any solution that passes through}\ S\ \text{will follow the}\)

\(\text{slope field (not crossing any lines) and approach a horizontal}\)

\(\text{asymptote at}\ P=2000\ \text{from the lower side}\ (P<2000).\)

ii.    
       

iii.  \(\dfrac{d P}{d t}=P(2000-P)\)

\(\text {Find \(P\) where  \(\dfrac{d P}{d t}\)  is a maximum.}\)

\(\text{Consider the graph}\ \ y=P(2000-P): \)

\(\Rightarrow \ \text {Graph is a concave down quadratic cutting at}\ \ P=0\ \ \text{and}\ \ P=2000\)

\(\Rightarrow \ \text{Max value of}\ \ P(2000-P)\ \ \Big(\text{i.e.}\ \dfrac{dP}{dt}\Big)\ \ \text{occurs at}\ \ P=1000\ \text{(axis).}\)

♦ Mean mark (iii) 51%.

CHEMISTRY, M2 EQ-Bank 14

The concentration of a solution of hydrochloric acid \(\ce{HCl}\) is 1.80% (w/v). What is the molar concentration produced by diluting 20.0 mL of this solution to a total volume of 200.0 mL with deionised water?   (3 marks) 

Show Answers Only

\(0.049\ \text{mol/L}\)

Show Worked Solution

→ 1.8% (w/v) means that there is 1.8 g of solute in 100 mL.

→ In 20 mL, there is \(1.8\ \text{g} \times 0.2 = 0.36\ \text{g}\) of \(\ce{HCl}\).

\(n\ce{(HCl)} = \dfrac{0.36}{1.008 + 35.45} = 0.00987\ \text{mol}\)

→ The final concentration after the dilution will be:

\(c = \dfrac{n}{V} = \dfrac{0.00987}{0.2} = 0.049\ \text{mol/L}\)

CHEMISTRY, M2 EQ-Bank 13

  1. Calculate the volume of solution needed to obtain 0.6 moles of solute from a solution of concentration 1.2 mol/L.   (1 mark)
  1. Explain the significance of accurate volume measurements when preparing solutions of specific concentrations in the laboratory.   (2 marks)
Show Answers Only

a.    \(0.5\ \text{L}\)

b.    The exact volumes of solutions must be known as:

→ The concentration of each solution must be precise, as it affects the molar ratios and yields.

→ Miscalculating volume would lead to incorrect concentrations, impacting experimental results and validity.

Show Worked Solution

a.    \(V=\dfrac{n}{c}=\dfrac{0.6}{1.2}=0.5\ \text{L}\)
 

b.    The exact volumes of solutions must be known as:

→ The concentration of each solution must be precise, as it affects the molar ratios and yields.

→ Miscalculating volume would lead to incorrect concentrations, impacting experimental results and validity.

Statistics, EXT1 S1 2024 HSC 9 MC

A bag contains \(n\) metal coins, \(n \ge 3\), that are made from either silver or bronze.

There are \(k\) silver coins in the bag and the rest are bronze.

Two coins are to be drawn at random from the bag, with the first coin drawn not being replaced before the second coin is drawn.

Which of the following expressions will give the probability that the two coins drawn are made of the same metal?
 

  1. \(\dfrac{k(k-1)+(n-k)(n-k-1)}{n(n-1)}\)
  2. \(\left(\begin{array}{l}n \\ 2\end{array}\right) \left(\begin{array}{l}n \\ k\end{array}\right) \left(\begin{array}{l}1-\dfrac{k}{n}  \end{array}\right)^{n-2} \)
  3. \(\dfrac{\left(\begin{array}{l}k \\ 2\end{array}\right) + \left(\begin{array}{c}n-k \\ 2\end{array}\right)}{n(n-1)}  \)
  4. \(\dfrac{k^{2}+(n-k)^2}{n^{2}}\)
Show Answers Only

\(A\)

Show Worked Solution

\(P(B)=P(\text{Bronze}), \ P(S)=P(\text{Silver}) \)

\(n=\ \text{total coins},\ k=\ \text{bronze coins},\ n-k=\ \text{silver coins}\)
 

\(P(BB \cup SS)\) \(=\dfrac{k}{n} \times \dfrac{k-1}{n-1} + \dfrac{n-k}{n} \times \dfrac{n-k-1}{n-1}\)  
  \(=\dfrac{k(k-1)+(n-k)(n-k-1)}{n(n-1)}\)  

 
\(\Rightarrow A\)

♦ Mean mark 54%.

CHEMISTRY, M2 EQ-Bank 11

A student is investigating the concentration of copper ions in a water sample collected from a local river. They use an instrument to determine that the sample contains copper ions at a concentration level of 1.75 ppm.

  1. Calculate the mass of copper ions in a 2 L sample of water.   (2 marks)
  1. Explain why parts per million is a suitable unit for measuring low concentrations of ions in environmental samples like river water.   (2 marks)
Show Answers Only

a.    3.5 mg

b.    Benefits of using ppm as a concentration measurement.

→ Parts per million (ppm) is suitable for measuring low concentrations of ions because it represents the mass of solute per million parts of solution, making it ideal for detecting trace levels of substances.

→ This unit allows for easy comparison of small concentrations, which is especially useful in environmental studies where contaminants are present in very low amounts.

Show Worked Solution

a.    \(1\ \text{ppm} = 1\ \text{mg/L}\)

\(\Rightarrow \ce{[Cu^{2+}]} = 1.75\ \text{mg/L}\)

\(m\ce{(Cu^{2+})} = 1.75 \times 2 = 3.5\ \text{mg}\)

 

b.    Benefits of using ppm as a concentration measurement.

→ Parts per million (ppm) is suitable for measuring low concentrations of ions because it represents the mass of solute per million parts of solution, making it ideal for detecting trace levels of substances.

→ This unit allows for easy comparison of small concentrations, which is especially useful in environmental studies where contaminants are present in very low amounts.

CHEMISTRY, M2 EQ-Bank 10

  1. A student is asked to prepare 500.0 mL of a 0.150 mol L\(^{-1}\) standard solution of oxalic acid \(\ce{(C2H2O4.2H2O)}\), and then to perform a dilution to produce 250.0 mL of a 0.0300 mol L\(^{-1}\) solution. Outline and explain each step in this process, including the calculations involved and choice of equipment.   (5 marks)
  1. Justify the procedure in part (a.) by explaining two measures taken to ensure the accuracy of the standard solution and diluted solution produced.   (2 marks)
Show Answers Only

a.    Calculate the mass of oxalic acid:

\(MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}\)

\(n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}\)

\(m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}\)

→ Measure 9.455 g of oxalic acid using an electronic balance.
 

Prepare the standard solution:

→ Transfer the oxalic acid to a 500.0 mL volumetric flask using a funnel and rinse any remaining crystals from the weighing container into the flask with a small amount of distilled water.

→ Fill the volumetric flask with distilled water to about 80% full and swirl to dissolve the oxalic acid completely and carefully add more distilled water with a pipette until the bottom of the meniscus rests on the 500.0 mL mark to ensure precise volume.

→ Stopper the flask and invert several times to ensure a homogeneous solution.
 

Perform the dilution:

\(V_1=\dfrac{c_2V_2}{c_1} = \dfrac{0.03 \times 250}{0.15} = 50\ \text{mL}\)

→ Use a pipette to transfer 50.0 mL of the 0.150 mol L\(^{-1}\) solution into a 250.0 mL volumetric flask.

→ Dilute with distilled water up to the 250.0 mL mark in the flask to achieve a concentration of 0.0300 mol L\(^{-1}\).

 

b.    Answers could include two of the following:

→ A volumetric flask is used for both the standard solution and the diluted solution, as it provides precise measurements for the final solution volume. This accuracy is essential for ensuring the concentration is exactly as calculated.

→ The primary solute is weighed on an electronic balance to the nearest 0.01 g or better, minimising any error in the amount of solute added to the solution. 

→ The pipette provides precise measurements crucial for accurate dilutions.

→ Using oxalic acid as the primary standard for the investigation. Primary standard’s have high molar masses and are anhydrous (don’t absorb water). This ensures the substance is pure and the electronic balance can accurately weigh the sample.

Show Worked Solution

a.    Calculate the mass of oxalic acid:

\(MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}\)

\(n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}\)

\(m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}\)

→ Measure 9.455 g of oxalic acid using an electronic balance.
 

Prepare the standard solution:

→ Transfer the oxalic acid to a 500.0 mL volumetric flask using a funnel and rinse any remaining crystals from the weighing container into the flask with a small amount of distilled water.

→ Fill the volumetric flask with distilled water to about 80% full and swirl to dissolve the oxalic acid completely and carefully add more distilled water with a pipette until the bottom of the meniscus rests on the 500.0 mL mark to ensure precise volume.

→ Stopper the flask and invert several times to ensure a homogeneous solution.
 

Perform the dilution:

\(V_1=\dfrac{c_2V_2}{c_1} = \dfrac{0.03 \times 250}{0.15} = 50\ \text{mL}\)

→ Use a pipette to transfer 50.0 mL of the 0.150 mol L\(^{-1}\) solution into a 250.0 mL volumetric flask.

→ Dilute with distilled water up to the 250.0 mL mark in the flask to achieve a concentration of 0.0300 mol L\(^{-1}\).

 

b.    Answers could include two of the following:

→ A volumetric flask is used for both the standard solution and the diluted solution, as it provides precise measurements for the final solution volume. This accuracy is essential for ensuring the concentration is exactly as calculated.

→ The primary solute is weighed on an electronic balance to the nearest 0.01 g or better, minimising any error in the amount of solute added to the solution. 

→ The pipette provides precise measurements crucial for accurate dilutions.

→ Using oxalic acid as the primary standard for the investigation. Primary standard’s have high molar masses and are anhydrous (don’t absorb water). This ensures the substance is pure and the electronic balance can accurately weigh the sample.

Statistics, EXT1 S1 2024 HSC 7 MC

A driver's knowledge test contains 30 multiple-choice questions, each with 4 options. An applicant must get at least 29 correct to pass.

If an applicant correctly answers the first 25 questions and randomly guesses the last 5 questions, what is the probability that the applicant will pass the test?

  1. \(\dfrac{1}{256}\)
  2. \(\dfrac{15}{1024}\)
  3. \(\dfrac{1}{64}\)
  4. \(\dfrac{21}{256}\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{For each of the last 5 questions:}\)

\(P(C) = \dfrac{1}{4}, \ \ P(\bar{C}) = \dfrac{3}{4}\)

\(P(\text{at least 4 correct})\) \(=\ ^5C_4 \Bigg(\dfrac{1}{4}\Bigg)^{4} \Bigg(\dfrac{3}{4}\Bigg)^{1} + \ ^5C_5 \Bigg(\dfrac{1}{4}\Bigg)^{5}\Bigg(\dfrac{3}{4}\Bigg)^{0}\)  
 

\(=5 \times \dfrac{1}{256} \times \dfrac{3}{4}+1\times \dfrac{1}{1024} \times 1\)

  
  \(=\dfrac{1}{64}\)  

 
\(\Rightarrow C\)

♦ Mean mark 44%.

Functions, EXT1 F1 2024 HSC 6 MC

How many real value(s) of \(x\) satisfy the equation

\(\abs{b} = \abs{b\,\sin(4x)}\),

where  \(x \in [0, 2\pi]\) and \(b\) is not zero?

  1. \(1\)
  2. \(2\)
  3. \(4\)
  4. \(8\)
Show Answers Only

\(D\)

Show Worked Solution

\(\abs{b} = \abs{b\,\sin(4x)}\ \ \Rightarrow\ \ \abs{1} = \abs{\sin(4x)}\ \ (b \neq 0)\)

\(\sin(4x) = \pm 1\)

\(x \in [0, 2\pi]\ \ \Rightarrow\ \ 4x \in [0, 8\pi]\)

\(\therefore 8\ \text{real solutions}\)

\(\Rightarrow D\)

♦ Mean mark 53%.

Statistics, 2ADV S3 2024 HSC 25

A function \(f(x)\) is defined as

\(f(x)=\left\{\begin{array}{ll} 0, & \text { for}\ \ x \lt 0 \\
1-\dfrac{x}{h}, & \text { for}\ \ 0 \leq x \leq h, \\
0, & \text { for}\ \  x \gt h \end{array}\right.\)

where \(h\) is a constant.

  1. Find the value of \(h\) such that \(f(x)\) is a probability density function.   (2 marks)

  2. By first finding a formula for the cumulative distribution function, sketch its graph.   (2 marks)

  3. Find the value of the median of the probability density function \(f(x)\) . Give your answer correct to 3 decimal places.   (2 marks)

Show Answers Only

a.   \(h=2\)

b.   

c.   \(\text{Median}\ =0.586\)

Show Worked Solution

a.   \(f(x)\ \text{is a PDF if:}\)

\(\displaystyle \int_{0}^{h} 1-\dfrac{x}{h}\,dx\) \(=1\)  
\(\Big[x-\dfrac{x^2}{2h} \Big]_0^{h}\) \(=1\)  
\(h-\dfrac{h}{2}\) \(=1\)  
\(h\) \(=2\)  

  
b.   \( \displaystyle \int 1-\dfrac{x}{2}\,dx = x-\dfrac{x^2}{4} + c\)

\(\text{At}\ \ x=0, \ \ F(0)=0,\ \ c=0 \)

\(f(x)=\left\{\begin{array}{ll} 0, & \text { for}\ \ x \lt 0 \\
x-\dfrac{x^2}{4}, & \text { for}\ \ 0 \leq x \leq 2, \\
1, & \text { for}\ \  x \gt 2 \end{array}\right.\)
 

♦♦ Mean mark (b) 27%.

c.   \(\text{Let}\ \ m=\ \text{median of}\ f(x) \)

\(\displaystyle \int_0^{m} 1-\dfrac{x}{2}\,dx\)  \(=0.5\)  
\(\Big[ x-\dfrac{x^2}{4} \Big]_0^m\) \(=0.5\)  
\(m-\dfrac{m^2}{4}\) \(=0.5\)  
\(4m-m^2\) \(=2\)  
\(m^2-4m+2\) \(=0\)  
\((m-2)^2\) \(=2\)  
\(m\) \(=2 \pm \sqrt{2}\)  

 

\(\therefore \ \text{Median}\) \(=2-\sqrt{2}\ \ (x \in [0,2]) \)   
  \(=0.586\ \text{(3 d.p.)}\)  
♦♦ Mean mark (c) 38%.

Measurement, STD1 M5 2024 HSC 32

A scale diagram is shown with locations \(A, B\) and \(C\) marked (assume grid squares are 1 cm × 1 cm).

Jo takes 24 minutes to walk from \(A\) to \(B\) (in a straight line) when walking at 3 km per hour.
 

  1. What is the scale used in the diagram?   (3 marks)
  2. What is the distance from \(B\) to \(C\), in kilometres?   (2 marks)
Show Answers Only

a.    \(1:20\,000\)

b.    \(\text{1.4 km}\)

Show Worked Solution

a.   \(\text{Find distance }A\rightarrow B:\)

\(D=S \times T = 3 \times \dfrac{24}{60} = 1.2\ \text{km}\)
  

  \(\text{Scale }\rightarrow\ 6\ \text{cm }\) \(:\ 1.2\ \text{km}\)
  \(6\ \text{cm }\) \(:\ 1200\ \text{m}\)
  \(6\ \text{cm }\) \(:\ 1200\times 100\ \text{cm}\)
  \(6\ \) \(:\ 120\,000\)
  \(1\ \) \(:\ 20\,000\)
♦♦♦ Mean mark 15%.

b.   \(\text{Distance }B\rightarrow C = 7\ \text{grid units}\)

\(\text{Real distance }B\rightarrow C\) \(=7 \times 20\,000\)
  \(=140\,000\ \text{cm}\)
  \(=1400\ \text{m}\)
  \(=1.4\ \text{km}\)
♦♦♦ Mean mark 28%.

Algebra, STD1 A3 2024 HSC 23

Carrie is organising a fundraiser.

The cost of hiring the venue and the band is $2500. The cost of providing meals is $50 per person.

  1. Complete the table of values to show the total cost of the fundraiser.   (1 mark)

\begin{array} {|l|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Number of people} \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \  & 25 & 50 & 75 & 100 & 125 & 150 \\
\hline
\rule{0pt}{2.5ex} \text{Cost} \rule[-1ex]{0pt}{0pt} & & 3750 & 5000 & 6250 & 7500 & 8750 & 10\,000 \\
\hline
\end{array}

  1. Carrie decides that tickets should be sold at $70 per person. The graph shows the expected revenue at this ticket price. Using the information in part (a), plot the line that shows the cost of the fundraiser.   (2 marks)

     

  1. How many tickets need to be sold for the fundraiser to break even?   (1 mark)

  2. Carrie sold 300 tickets. How much profit did the fundraiser make?   (3 marks)

Show Answers Only

a.    \(\text{Cost when 0 people }= $2500\)

b.


c.    \(100\ \text{tickets}\)

d.    \(\text{Profit }=$3500\)

Show Worked Solution

a.    \(\text{Cost when 0 people }= $2500\)

b.


c.    \(\text{Point of intersection}\ \ \Rightarrow\ \ \text{break-even}\)

\(\therefore\ \text{Break-even when 100 tickets sold.}\)
 

Mean mark (c) 53%.

d.    \(\text{Revenue}\ (R)=70n\ \ (n=\ \text{number of people)}\)

\(\text{Cost}\ (C)=2500 + \Big(\dfrac{1250}{25}\Big)n=2500+50n\)

\(\text{Find profit}\ (P)\ \text{when}\ \ n=300:\)

\(P\) \(=R-C\)
  \(=70\times 300-(2500+50\times300)\)
  \(=21\,000-17\,500\)
  \(=$3500\)
♦ Mean mark (d) 46%.

Measurement, STD1 M5 2024 HSC 29

A floor plan for a living area is shown. All measurements are in millimetres.
 

  1. What is the length and width of the cupboard, in metres?   (1 mark)

  2. The floor of the living area is to be tiled. Tiles will NOT be placed under the cupboard.
  3. Each tile is 0.2 m × 0.5 m. The tiles are supplied in boxes of 15 at a cost of $100 per box. Only full boxes can be purchased.
  4. What is the cost of the tiles for the living area?   (4 marks)
Show Answers Only

a.    \(4\ \text{m}\times 0.5\ \text{m}\)

b.    \($1100\)

Show Worked Solution

a.    \(\text{1 metre = 1000 mm}\)

\(\text{Cupboard}\ = 4\ \text{m}\times 0.5\ \text{m}\)

Mean mark (a) 53%.

b.    \(\text{Method 1}\)

\(\text{Total area to be tiled}\ =6\times 3-4\times 0.5=16\ \text{m}^2\)

\(\text{Area of each tile}\ =0.2\times0.5=0.1\ \text{m}^2\)

\(\text{Tiles needed}\ =\dfrac{16}{0.1}=160\ \text{tiles}\)

\(\text{Number of boxes}\ =\dfrac{160}{15}=10.66\dots=11\ \text{boxes}\)

\(\therefore\ \text{Total cost}\ =100\times 11=$1100\)

   
\(\text{Method 2}\)

\(\text{Tiles to fit 6 m width}\ =\dfrac{6}{0.2}=30 \)

\(\text{Tiles to fit 2 m width}\ =\dfrac{2}{0.2}=10 \)
 

\(\text{Total tiles}\ =30\times 5\ \text{rows}+10\times 1\ \text{rows}\ =160\)

\(\text{Number of boxes}\ =\dfrac{160}{15}=10.66\dots=11\ \text{boxes}\)

\(\therefore\ \text{Total cost}\ =100\times 11=$1100\)

♦♦ Mean mark (b) 35%.