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Calculus, EXT1 C3 2025 HSC 12d

Find the solution of  \(\dfrac{dy}{dx}=\sqrt{(2-y)(2+y)}\), given that  \(y=1\)  when  \(x=0\).   (3 marks)

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\(y=2\, \sin \left(x+\dfrac{\pi}{6}\right)\)

Show Worked Solution
\(\dfrac{d y}{d x}\) \(=\sqrt{(2-y)(2+y)}\) \(=\sqrt{4-y^2}\)
\(\dfrac{d x}{d y}\) \(=\dfrac{1}{\sqrt{4-y^2}}\)
\(\displaystyle \int d x\) \(=\displaystyle \int \dfrac{1}{\sqrt{4-y^2}} d y\)
\(x\) \(=\sin ^{-1}\left(\dfrac{y}{2}\right)+c\)

 

\(\text{When} \ \ x=0, y=1:\)

\(0=\sin ^{-1}\left(\dfrac{1}{2}\right)+c \ \ \Rightarrow \ \ c=-\dfrac{\pi}{6}\)

\(x\) \(=\sin ^{-1}\left(\dfrac{y}{2}\right)-\dfrac{\pi}{6}\)
\(\sin ^{-1}\left(\dfrac{y}{2}\right)\) \(=x+\dfrac{\pi}{6}\)
\(\dfrac{y}{2}\) \(=\sin \left(x+\dfrac{\pi}{6}\right)\)
\(y\) \(=2\, \sin \left(x+\dfrac{\pi}{6}\right)\)

Proof, EXT1 P1 2025 HSC 12c

Prove by mathematical induction that

\(1 \times(1!)+2 \times(2!)+\cdots+n \times(n!)=(n+1)!-1\)

for integers  \(n \geq 1\).   (3 marks)

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\(\text{See Worked Solution}\)

Show Worked Solution

\(1 \times 1!+2 \times 2!+\ldots+n \times n!=(n+1)!-1\)

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

\(\text{LHS}=1 \times 1!=1\)

\(\text{RHS}=(2)!-1=1=\text{LHS }\)

\(\therefore\ \text{True for}\ \ n=1.\)
 

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

\(1 \times 1!+2 \times 2!+\ldots+k \times k!=(k+1)!-1\)
 

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

\(\text{i.e.} \ \ 1 \times 1!+2 \times 2!+\ldots +k \times k!+(k+1) \times(k+1)!=(k+2)!-1\)

\(\text{LHS}\) \(=1\times 1!+\ldots+k \times k!+(k+1) \times(k+1)!\)
  \(=(k+1)!-1+(k+1) \times (k+1)!\)
  \(=(k+1)!(1+k+1)-1\)
  \(=(k+1)!(k+2)-1\)
  \(=(k+2)!-1\)
  \(= \operatorname{RHS}\)

 

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

\(\therefore \ \text{Since true for  \(\ n=1\),  by PMI, true for integers  \(\ n \geqslant 1\).}\)

Calculus, EXT1 C3 2025 HSC 12b

Consider the region bounded by the hyperbola  \(y=\dfrac{1}{x}\),  the \(y\)-axis and the lines  \(y=1\)  and  \(y=a\)  for  \(a>1\).

Find the volume of the solid of revolution formed when the region is rotated about the \(y\)-axis.   (2 marks)

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\(\pi\left(1-\dfrac{1}{a}\right)\ \text{u}^3\)

Show Worked Solution

\(y=\dfrac{1}{x} \ \Rightarrow \ x^2=\dfrac{1}{y^2}\)

\(V\) \(=\pi \displaystyle \int_1^a x^2\, dy\)
  \(=\pi \displaystyle \int_1^a y^{-2}\, d y\)
  \(=-\pi\left[y^{-1}\right]_1^a\)
  \(=-\pi\left(\dfrac{1}{a}-1\right)\)
  \(=\pi\left(1-\dfrac{1}{a}\right)\ \text{u}^3\)

Calculus, EXT1 C1 2025 HSC 12a

The radius, \(r\) cm, and angle, \(\theta\) radians, of a sector vary in such a way that its area remains a constant 10 cm².
 

The angle \(\theta\) is increasing at a constant rate of 2 radians per second.

Find the rate at which the radius is changing when the radius is 4 cm.   (3 marks)

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\(\text{Radius is decreasing at} \ \dfrac{16}{5} \ \text{cm per second.}\)

Show Worked Solution

\(\dfrac{d \theta}{d t}=2\)

\(\dfrac{dv}{d t}=\dfrac{dv}{d \theta} \times \dfrac{d \theta}{d t}\)

\(A\) \(=\dfrac{\theta}{2 \pi} \times \pi r^2\)  
\(10\) \(=\dfrac{\theta}{2} \times r^2\)  
\(\theta\) \(=20\, r^{-2}\)  

 
\(\dfrac{d \theta}{dv}=-40 r^{-3} \)

\(\dfrac{dv}{d \theta}=-\dfrac{r^3}{40}\)
 

\(\text{ Find \(\dfrac{dv}{dt}\)  when  \(r=4\):}\)

\(\dfrac{dv}{dt}=-\dfrac{4^3}{40} \times 2=-\dfrac{16}{5}\)

\(\therefore \ \text{Radius is decreasing at} \ \dfrac{16}{5} \ \text{cm per second.}\)

Calculus, EXT1 C2 2025 HSC 11g

Evaluate \(\displaystyle\int_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}} \cos ^2(3 x) d x\).   (3 marks)

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\(\dfrac{\pi}{12}\)

Show Worked Solution
\(\displaystyle\int_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}} \cos ^2(3x) dx\) \(=\displaystyle \dfrac{1}{2} \int_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}} (\cos6x+1) dx \)
  \(=\dfrac{1}{2}\left[\dfrac{1}{6} \sin 6 x+x\right]_{\small{\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{3}}}\)
  \(=\dfrac{1}{2}\left[\left(\dfrac{1}{6} \sin 2 \pi+\dfrac{\pi}{3}\right)-\left(\dfrac{1}{6} \sin \pi+\dfrac{\pi}{6}\right)\right]\)
  \(=\dfrac{1}{2}\left(\dfrac{\pi}{3}-\dfrac{\pi}{6}\right)\)
  \(=\dfrac{\pi}{12}\)

Trigonometry, EXT1 T1 2025 HSC 11d

Sketch the graph of  \(y=\dfrac{1}{3} \cos ^{-1}(2 x)\).   (2 marks)

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Show Worked Solution

\(y=\dfrac{1}{3} \cos ^{-1}(2 x)\)

\(\text{Domain:} \ -1 \leqslant 2 x \leqslant 1 \ \Rightarrow \ -\dfrac{1}{2} \leqslant x \leqslant \dfrac{1}{2}\)

\(\text{Range:} \ \cos (2 x) \in[0, \pi] \  \Rightarrow \  \dfrac{1}{3} \cos (2x) \in\left[0, \dfrac{\pi}{3}\right]\)
 

Calculus, EXT1 C2 2025 HSC 11c

Find \(\displaystyle \int \sin 3x \, \cos x \, dx\).   (2 marks)

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\(-\dfrac{1}{8} \cos 4 x-\dfrac{1}{4} \cos 2 x+c\)

Show Worked Solution
\(\displaystyle\int \sin3x \, \cos x \, dx\) \(=\displaystyle\frac{1}{2} \int \sin 4 x+\sin2x \,dx\)
  \(=-\dfrac{1}{8} \cos 4 x-\dfrac{1}{4} \cos 2 x+c\)

Calculus, EXT1 C2 2025 HSC 3 MC

Consider the integral  \(\large{\displaystyle{\int}}_{\small{-\dfrac{5}{2}}}^{\small{\dfrac{5}{2}}}\) \(\left(\dfrac{1}{25-x^2}\right) d x\).

The substitution  \(x=5 \sin \theta\)  is applied.

Which of the following is obtained?

  1. \(\dfrac{1}{5}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\operatorname{cosec} \theta \, d \theta\)
  2. \(\dfrac{1}{5}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\sec \theta \,  d \theta\)
  3. \(\dfrac{1}{25}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\operatorname{cosec}^2 \theta \, d \theta\)
  4. \(\dfrac{1}{25}\large{\displaystyle{\int}}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\sec ^2 \theta \, d \theta\)
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\(\Rightarrow B\)

Show Worked Solution

\(x=5\, \sin \theta\)

\(\dfrac{dx}{d \theta}=5\, \cos \theta \ \Rightarrow \ dx=5\, \cos \theta \, d \theta\)

\(\text{When} \ \ x=\dfrac{5}{2} \ \Rightarrow \ \sin\, \theta=\dfrac{1}{2} \ \Rightarrow \ \theta=\dfrac{\pi}{6}\)

\(\text{When} \ \ x=-\dfrac{5}{2} \ \Rightarrow \ \sin \theta=-\dfrac{1}{2} \ \Rightarrow \ \theta=-\dfrac{\pi}{6}\)

\(\large{\displaystyle{\int}}_{\small{-\dfrac{5}{2}}}^{\small{\dfrac{5}{2}}}\) \(\left(\dfrac{1}{25-x^2}\right) d x\) \(=\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\left(\dfrac{1}{25-25\, \sin ^2 \theta}\right) \cdot 5\  \cos \theta \, d \theta\)
  \(=\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\dfrac{5\, \cos \theta}{25\, \cos ^2 \theta} \,  d \theta\)
  \(=\dfrac{1}{5}\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\dfrac{1}{\cos \theta} \, d \theta\)
  \(=\dfrac{1}{5}\large{\displaystyle \int}_{\small{-\dfrac{\pi}{6}}}^{\small{\dfrac{\pi}{6}}}\)\(\sec \theta \, d \theta\)

\(\Rightarrow B\)

Calculus, EXT1 C3 2025 HSC 7 MC

A slope field is shown.
 

Which of the following could be the differential equation represented by the slope field?

  1. \(\dfrac{d y}{d x}=x^2\)
  2. \(\dfrac{d y}{d x}=x^2+C, C \neq 0\)
  3. \(\dfrac{d y}{d x}=x^3\)
  4. \(\dfrac{d y}{d x}=x^3+C, C \neq 0\)
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\(A\)

Show Worked Solution

\(\text{For all \(x<0\), gradients are positive (from graph):}\)

\(\text{Eliminate C and D.}\)

\(\text{At \(x=0\), gradient = 0 (from graph):}\)

\(\text{Eliminate B.}\)

\(\Rightarrow A\)

Calculus, EXT1 C3 2025 HSC 6 MC

Given that \(a\) is a non-zero constant, which of the following integrals is equal to zero?

  1. \(\displaystyle \int_{-a}^a x\, \cos ^{-1}(x) d x\)
  2. \(\displaystyle\int_{-a}^a x^2\, \cos ^{-1}(x) d x\)
  3. \(\displaystyle\int_{-a}^a x\, \tan ^{-1}(x) d x\)
  4. \(\displaystyle\int_{-a}^a x^2\, \tan ^{-1}(x) d x\)
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\(D\)

Show Worked Solution

\(\text{Since the limits are symmetrical about 0:}\)

\(\text{Integral will equal zero if function is odd.}\)

\(\text{Consider option D:}\)

\(f(x)=x^2\, \tan^{-1}(x)\)

\(f(-x)=(-x)^2\, \tan^{-1}(-x)=-x^2\, \tan ^{-1}(x)=-f(x)\ \text{(odd)}\)

\(\Rightarrow D\)

Statistics, EXT1 S1 2025 HSC 4 MC

A Bernoulli random variable \(X\) has probability distribution

\(P(x)=\dfrac{x+1}{3}\)  for  \(x=0,1\).

What are the mean and variance of \(X\) ?

  1. \(E(X)=\dfrac{1}{3}, \quad \operatorname{Var}(X)=\dfrac{2}{9}\)
  2. \(E(X)=\dfrac{1}{3}, \quad \operatorname{Var}(X)=\dfrac{2}{3}\)
  3. \(E(X)=\dfrac{2}{3}, \quad \operatorname{Var}(X)=\dfrac{2}{9}\)
  4. \(E(X)=\dfrac{2}{3}, \quad \operatorname{Var}(X)=\dfrac{2}{3}\)
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\(C\)

Show Worked Solution

\(P(0)=\dfrac{1}{3}, \ P(1)=\dfrac{2}{3} \)

\(E(X) = \dfrac{1}{3} \times 0 + \dfrac{2}{3} \times 1 = \dfrac{2}{3}\)

\(E(X^2) = \dfrac{1}{3} \times 0^2 + \dfrac{2}{3} \times 1^2 = \dfrac{2}{3} \)

\(\text{Var}(X) = E(X^2)-E(X)^2 = \dfrac{2}{3}-\dfrac{4}{9}=\dfrac{2}{9} \)

\(\Rightarrow C\)

Trigonometry, 2ADV T3 2025 HSC 15

A sound wave can be modelled using a function  \(P(t)=k\, \sin a t\), where \(P\) is air pressure in Pascals, \(t\) is time in milliseconds (ms) and \(k\) and \(a\) are constants.

  1. Write the equation for a sound wave \(P_1(t)\) that has an amplitude of 2 Pascals and a period of 5 ms.   (2 marks)

  2. The graph of \(P_1(t)\) from part (a) is shown.
  3. On the diagram, sketch the graph of  \(P_2(t)=4 \sin \left(\dfrac{\pi}{10} t\right)\)  for  \(0 \leq t \leq 10\).   (2 marks)
     

  1. Hence, find the values of \(t\), where  \(0<t<10\),  for which functions \(P_1(t)\) and \(P_2(t)\) are BOTH decreasing.   (2 marks)

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a.  \(P_1(t)=2\, \sin \left(\dfrac{2 \pi t}{5}\right)\)

b.   
       

c.  \(\text{Both decreasing for} \ \ 6.25<t<8.75\)

Show Worked Solution

a.    \(\text{Amplitude}=2 \Rightarrow k=2\)

\(\text{Period}=5\)

\(\dfrac{2 \pi}{a}\) \(=5\)
\(5a\) \(=2 \pi\)
\(a\) \(=\dfrac{2 \pi}{5}\)

 
\(\therefore P_1(t)=2\, \sin \left(\dfrac{2 \pi t}{5}\right)\)
 

b.   
     

\(P_2(t)=4\, \sin \left(\dfrac{\pi}{10} t\right)\)

\(\text{Amplitude}=4\)

\(\text{Period}=\dfrac{2 \pi}{\frac{\pi}{10}}=20\)
 

c.    \(\text {By inspection of graph:}\)

\(P_2(t) \ \text {is decreasing for} \ \ 5<t \leq10\)

\(P_1(t) \text { is decreasing for} \ \ 1.25<t<3.75 \ \ \text{and}\ \ 6.25<t<8.75\)

\(\therefore \ \text{Both decreasing for} \ \ 6.25<t<8.75\)

Functions, 2ADV F2 2025 HSC 6 MC

The graph of  \(y=f(x)\) is shown.  
 

Which of the following is the graph of   \(y=-f(-x)\) ?
 

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\(C\)

Show Worked Solution

\(y=-f(x)\ \ \Rightarrow\ \ \text{Reflect \(f(x)\) in the \(x\)-axis.}\)

\(y=-f(-x)\ \ \Rightarrow\ \ \text{Reflect \(-f(x)\) in the \(y\)-axis.}\)

\(\text{In this combination of translations, the order is not important.}\)

\(\Rightarrow C\)

Networks, STD2 N3 2025 HSC 22

A network of pipes with one cut is shown. The number on each edge gives the capacity of that pipe in L/min.
 

  1. What is the capacity of the cut shown?   (1 mark)

  2. The diagram shows a possible flow for this network of pipes.
     

    1. What is the value of \(x\)? Give a reason for your answer.   (2 marks)

    2. Which of the pipes in the flow are at full capacity?   (1 mark)

    3. The maximum flow for this network is 50 L/min.
    4. Which path of pipes could have an increase in flow of 2 L/min to achieve the maximum flow?   (1 mark)

Show Answers Only

a.    \(\text{Capacity} =62\)

b.i.   \(x=30\) 

b.ii. \(DE, DG, CF \ \text{and} \ FG\)

b.iii.  \(ACEG\)

Show Worked Solution

a.    \(\text{Capacity} =26+24+12=62\)
 

b.i.   \(\text{Inflow into} \ C\) \(=\text{Outflow from} \ C\)
  \(x\) \(=5+13+12\)
    \(=30\)

 

b.ii. \(DE, DG, CF \ \text{and} \ FG\)
 

b.iii.  \(ACEG\)

Statistics, STD2 S4 2025 HSC 25

In a research study, participants were asked to record the number of minutes they spent watching television and the number of minutes they spent exercising each day over a period of 3 months. The averages for each participant were recorded and graphed.
 

 

  1. Describe the bivariate dataset in terms of its form and direction.   (2 marks)
  2. Form:  ..................................................................
  3. Direction:  ............................................................

The equation of the least-squares regression line for this dataset is

\(y=64.3-0.7 x\)

  1. Interpret the values of the slope and \(y\)-intercept of the regression line in the context of this dataset.   (2 marks)

  2. Jo spends an average of 42 minutes per day watching television.
  3. Use the equation of the regression line to determine how many minutes on average Jo is expected to exercise each day.   (1 mark)

  4. Explain why it is NOT appropriate to extrapolate the regression line to predict the average number of minutes of exercise per day for someone who watches an average of 2 hours of television per day.   (1 mark)

Show Answers Only

a.    \(\text{Form: Linear. Direction: Negative}\)

Show Worked Solution

a.    \(\text{Form: Linear}\)

\(\text{Direction: Negative}\)
 

b.    \(\text{Slope}=-0.7\)

\(\text{This means that for each added minute of watching television per day, a participant, on average,}\)

\(\text{will exercise for 0.7 minutes less.}\)

\(y \text{-intercept}=64.3\)

\(\text{If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.}\)
 

c.    \(\text{At} \ \ x=42:\)

\(y=64.3-0.7 \times 42=34.9\)

\(\therefore \ \text{Jo is expected to exercise for 34.9 minutes}\)
 

d.    \(\text{At} \ \  x=120\ \text{(2 hours),} \ \ y=64.3-0.7 \times 120=-19.7\)

\(\text{The model predicts a negative value for time spent exercising, which is not possible.}\)

Measurement, STD2 M7 2025 HSC 21

A house has a reverse-cycle air conditioner which uses 2.5 kW of power for cooling and 3.2 kW of power for heating. The cost of electricity is 29 cents per kWh .

  1. Find the cost, in dollars and cents, of cooling the house for 6 hours.   (1 mark)

  2. The cost of operating the air conditioner to heat the house during winter last year was $640. There are 92 days in winter.
  3. Find the number of hours, to 1 decimal place, that the air conditioner was used on average per day.   (2 marks)

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a.    \(\text{Cooling cost (6 hours )}\ = 6 \times 2.5 \times 0.29 = \$4.35\)

b.    \(7.5\ \text{hours}\)

Show Worked Solution

a.    \(\text{Cooling cost (6 hours )}\ = 6 \times 2.5 \times 0.29 = \$4.35\)
 

b.    \(\text{Let \(h\) = hours used per day}\)

\(\text{Cost per day}\ = h \times 3.2 \times 0.29\)

\(\text{Cost (92 days )}\ = 92 \times h \times 3.2 \times 0.29\)

\(\text{Find \(h\) when cost = \$640:}\)

\(640\) \(=92 \times h \times 3.2 \times 0.29\)  
\(h\) \(=\dfrac{640}{92 \times 3.2 \times 0.29}=7.5\ \text{hours (1 d.p.)}\)  

Networks, STD2 N3 2025 HSC 19

The activities and corresponding durations in days for a project are shown in the network diagram.
 

 

  1. Complete the table showing the immediate prerequisites for each activity. Indicate with an \(\text{X}\) any activities without any immediate prerequisites.   (2 marks)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Activity} \rule[-1ex]{0pt}{0pt} & \text{Immediate prerequisite(s)} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  \\
\hline
\end{array}

  1. Find the critical path for this project AND state the minimum duration for the project.   (2 marks)

  1. The duration of activity \( A \) is increased by 2. Does this affect the critical path for the project? Give a reason for your answer.   (1 mark)

Show Answers Only

a.           

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Activity} \rule[-1ex]{0pt}{0pt} & \text{Immediate prerequisite(s)} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & \text{X} \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & C,D \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & E \\
\hline
\end{array}

  
b.   \(\text{Critical Path:}\ BDEFH\)

\(\text{Minimum Duration}\ =4+5+5+7+5=26\ \text{days}\)
 

c.   \(\text{If duration of activity \(A\) is increased by 2:}\)

\(\text{The critical path remains unchanged (EST of activity \(E\) remains = 9)}\)

Show Worked Solution

a.   

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Activity} \rule[-1ex]{0pt}{0pt} & \text{Immediate prerequisite(s)} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & \text{X} \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & C,D \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & E \\
\hline
\end{array}

  
b.   

\(\text{Critical Path:}\ BDEFH\)

\(\text{Minimum Duration}\ =4+5+5+7+5=26\ \text{days}\)
 

c.   \(\text{If duration of activity \(A\) is increased by 2:}\)

\(\text{The critical path remains unchanged (EST of activity \(E\) remains = 9)}\)

Financial Maths, STD2 F5 2025 HSC 18

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

The prize in a lottery is an annuity of $5000 a year for 10 years, invested at 4.5% per annum compounding annually.

What will be the value of the prize at the end of 10 years?   (2 marks)

Show Answers Only

\($61\,440\)

Show Worked Solution

\( r=4.5\%\ \text{annually}\)

\(\text{Compounding periods = 10}\)

\(\text{Annuity factor = 12.288}\)

\(\therefore\ \text{FV (annuity)}\ = 5000 \times 12.288=$61\,440\)

Statistics, STD2 S4 2025 HSC 17

The scatter plot shows a bivariate dataset, where \(x\) is the independent variable and \(y\) is the dependent variable. 
 

 

The points \( (0,14) \) and \( (5,4)\) lie on the line of best fit.

Plot the points \( (0,14) \) and \( (5,4) \) on the graph and hence find the equation of the line of best fit.   (3 marks)

Show Answers Only

\( y=-2 x+14 \)

Show Worked Solution

\(\text{Gradient (LOBF)}\ =\dfrac{y_2-y_1}{x_4-x_1}=\dfrac{4-14}{5-0}=-2 \)

\(\text{Find equation of line}\ \ m=-2 \ \ \text{through}\ (0,14):\)

\( y-y_1\) \(=m(x-x_1)\)  
\( y-14\) \(=-2(x-0)\)  
\( y\) \(=-2 x+14 \)  

Measurement, STD2 M2 2025 HSC 12 MC

A football game is being played in Quito (UTC –5) starting at 3:40 pm on Tuesday.

What is the time in Sydney (UTC +10) when the game starts in Quito?

  1. 10:40 am Tuesday
  2. 8:40 pm Tuesday
  3. 12:40 am Wednesday
  4. 6:40 am Wednesday
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Sydney is 15 hours ahead of Quito.}\)

\(\text{3:40 pm Tuesday + 15 hours}\)

\(\text{= 3:40 am Wednesday + 3 hours}\)

\(\text{= 6:40 am Wednesday (Sydney time)}\)

\(\Rightarrow D\)

Measurement, STD2 M7 2025 HSC 10 MC

An electricity company charges customers 37 cents per kWh for electricity used, \( U\), and pays customers 5 cents per kWh for electricity produced, \(P\).

The electricity company also charges customers a fee of 71 cents per day.

Which formula should be used to calculate a customer's daily cost of electricity, \(C\), in cents?

  1. \( C=71+37 U-5 P \)
  2. \( C=71+37 U+5 P \)
  3. \( C=71-37 U-5 P \)
  4. \( C=71-37 U+5 P \)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Customer pays:}\ 71 + 37 \times U\)

\(\text{Customer receives:}\ 5 \times P\)

\(\text{Daily cost}\ = 71 + 37 \times U-5 \times P\)

\(\Rightarrow A\)

Measurement, STD2 M7 2025 HSC 9 MC

The ratio of the dimensions of a model car to the dimensions of an actual car is \(1:64\). The actual car has a length of 4.9 m.

What is the length of the model car in cm, correct to 1 decimal place?

  1. 3.1
  2. 7.7
  3. 13.1
  4. 59.1
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Actual length}=4.9\ \text{m}\ =490\ \text{cm}\)

\(\therefore\ \text{Model car length}\ =490\times\dfrac{1}{64}=7.65625\approx 7.7\ \text{cm}\)

 \(\Rightarrow B\)

Probability, STD2 S2 2025 HSC 8 MC

A spinner made up of 4 colours is spun 100 times. The frequency histogram shows the results.
 

Which of these spinners is most likely to give the results shown?
 

Show Answers Only

\(A\)

Show Worked Solution
\(P(\text{White})\) \(=\dfrac{50}{100}=\dfrac{1}{2}\)
\(P(\text{Red})\) \(=\dfrac{25}{100}=\dfrac{1}{4}\)  
\(P(\text{Yellow})\) \(=\dfrac{15}{100}=\dfrac{3}{20}\)
\(P(\text{Green})\) \(=\dfrac{10}{100}=\dfrac{2}{20}=\dfrac{1}{10}\)

 
\(\text{Eliminate Options B and D as white}\ \neq \dfrac{1}{2}\ \text{of spinner.}\)

\(\text{Eliminate Option C as red}\ \neq \dfrac{1}{4}\ \text{of spinner.}\)

\(\Rightarrow A\)

Networks, STD1 N1 2025 HSC 6 MC

The network shows the distances, in kilometres, along a series of roads that connect towns.
 

What is the value of the largest weighted edge included in the minimum spanning tree for this network?

  1. 7
  2. 8
  3. 9
  4. 10
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Minimum spanning tree:}\)
 

\(\text{Using Kruskal’s algorithm:}\)

\(\text{Edge 1 = 4, Edge 2/3 = 4, Edge 4 = 6, Edge 5 = 9}\) 

\(\therefore\ \text{The largest weighted edge in the MST = 9.}\)

\(\Rightarrow C\)

Calculus, 2ADV C4 2011 HSC 4d v1

  1. Differentiate  `y=sqrt(16 -x^2)`  with respect to  `x`.   (2 marks)

  2. Hence, or otherwise, find  `int (8x)/sqrt(16 -x^2)\ dx`.    (2 marks)

Show Answers Only
  1. `- x/sqrt(16\ -x^2)`
  2. `-8 sqrt(16\ -x^2) + C`
Show Worked Solution
IMPORTANT: Some students might find calculations easier by rewriting the equation as `y=(16-x^2)^(1/2)`.
a.    `y` `= sqrt(16 -x^2)`
    `= (16 -x^2)^(1/2)`

 

`dy/dx` `=1/2 xx (16 -x^2)^(-1/2) xx d/dx (16 -x^2)`
  `= 1/2 xx (16 -x^2)^(-1/2) xx -2x`
  `= – x/sqrt(16 -x^2)`

 

b.    `int (8x)/sqrt(16 – x^2)\ dx` `= -8 int (-x)/sqrt(16 -x^2)\ dx`
    `= -8 (sqrt(16 -x^2)) + C`
    `= -8 sqrt(16 -x^2) + C`

Calculus, 2ADV C4 EO-Bank 11

  1. Differeniate \(y=\dfrac{x}{x^2+1}\)  (2 marks)
  1. Hence evaluate \(\displaystyle \int_0^1{\dfrac{1-x^2}{(x^2+1)^2}}\, dx\)   (2 marks)
Show Answers Only

a.    \(\dfrac{dy}{dx} = \dfrac{1-x^2}{(x^2+1)^2}\)

b.    \(\dfrac{1}{2}\)

Show Worked Solution

a.    Using the quotient rule:

\(\dfrac{dy}{dx} = \dfrac{x^2+1-2x^2}{(x^2+1)^2} = \dfrac{1-x^2}{(x^2+1)^2}\)
 

b.    Using part a: 

 \(\displaystyle \int_0^1{\dfrac{1-x^2}{(x^2+1)^2}}\, dx\) \(=\left[\dfrac{x}{x^2+1}\right]_0^1\)  
  \(=\dfrac{1}{2} -0\)  
  \(=\dfrac{1}{2}\)  

 

Calculus, 2ADV C4 EO-Bank 10

Given that  `int_0^k ( 2x + 4 )\ dx = 21`, and  `k`  is a constant, find the value of  `k`.   (2 marks)

Show Answers Only

`k = 3`

Show Worked Solution
`int_0^k ( 2x + 4 ) \ dx` `= 21`
`int_0^k ( 2x + 4 ) \ dx` `= [ x^2 + 4x ]_0^k`
  `= [(k^2 + 4k ) – 0 ]`
  `= k^2 + 4k`

 

`=> k^2 + 4k` `=21`
`k^2+4k-21` `= 0`
`(k-3)(k+7)` `= 0`
`k` `=3, -7`
`k` `=3\ text(as ) k >0`

Calculus, 2ADV C1 2014 HSC 13c v1

The displacement of a particle moving along the  `x`-axis is given by

 `x =2t -3/sqrt(t+1)`,

where  `x`  is the displacement from the origin in metres,  `t`  is the time in seconds, and  `t >= 0`.

  1. Show that the acceleration of the particle is always negative.    (2 marks)

  2. What value does the velocity approach as  `t`  increases indefinitely?    (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `2`
Show Worked Solution
a.    `x` `=2t -3/sqrt(t+1)`
    `=2t -3(t+1)^(-1/2)`

 

`dot x` `= 2\ -3(-1/2) (t+1)^(-3/2)`
  `= 2 + 3/(2(t+1)^(3/2))`

 

`ddot x` `= -(9/4)(t+1)^(-5/2)`
  `= – 9/(4sqrt((t+1)^5))`

 
`text(S)text(ince)\ \ t >= 0,`

`=> 1/sqrt((t+1)^5) > 0`

`=> – 9/(4sqrt((t+1)^5)) < 0`
 

`:.\ text(Acceleration is always negative.)`

 

b.    `text(Velocity)\ (dot x) = 2 + 3/(2(t+1)^(3/2))`

 
`text(As)\ t -> oo,\ 3/(2(t+1)^(3/2)) -> 0`

`:.\ text(As)\ t -> oo,\ dot x -> 2`

Calculus, 2ADV C1 EO-Bank 11

A particle is moving along the `x`-axis. Its velocity `v` at time `t` is given by

`v = (t^2+4)/sqrt(3t+1)`  metres per second

Find the acceleration of the particle when  `t = 2`.

Express your answer as an exact value in its simplest form.  (3 marks)

Show Answers Only

` (16sqrt(7))/49\ \ text(ms)^(−2)`

Show Worked Solution

`v = (t^2+4)/sqrt(3t+1)`

`alpha` `= (dv)/(dt)`

`text(Using quotient rule:)`

`u=t^2+4,`     `v=(3t+1)^(1/2)`  
`u^{′} = 2t,`     `v^{′} = 3/2 (3t+1)^(-1/2)`  
     
`alpha` `= (u^{′} v-v^{′} u)/v^2`
  `= (2t (3t+1)^(1/2)-3/2(t^2+4) (3t+1)^(-1/2))/(3t+1)`

 
`text(When)\ \ t = 2,`

`alpha` `= (4(7)^(1/2)-12(7)^(-1/2))/(7)`
  `= (4sqrt(7))/7 -12/(7sqrt(7))`
  `= (28sqrt(7))/49-(12sqrt(7))/49`
  `= (16sqrt(7))/49\ \ text(ms)^(−2)`

Calculus, 2ADV C1 EO-Bank 3

The displacement `x` metres from the origin at time `t` seconds of a particle travelling in a straight line is given by

`x = t^3-4t^2 +5t + 6`     when   `t >= 0`

  1.  Calculate the velocity when  `t = 3`.  (1 mark)

  2.  When is the particle stationary?  (2 marks)

Show Answers Only

i.    `8\ text(ms)^(−1)`

ii.   `1, 5/3\ text(s)`

Show Worked Solution

i.   `x =t^3-4t^2 +5t + 6` 

`v = (dx)/(dt) = 3t^2-8t +5`

 
`text(When)\ t = 3,`

`v` `= 3 xx 3^2-8 · 3 +5`
  `= 8\ text(ms)^(−1)`

 

ii.   `text(Particle is stationary when)\ \ v = 0`

`3t^2-8t +5` `= 0`
`3t^2-3t-5t+5` `= 0`
`3t(t-1)-5(t-1)` `= 0`
`(t-1)(3t-5)` `= 0`
`t` `= 1, 5/3\ text(s)`

Functions, 2ADV F2 2025 HSC 30

The parabola with equation  \(y=(x-1)(x-5)\)  is translated both horizontally to the right and vertically up by \(k\) units, where \(k\) is positive.

The translated parabola passes through the point \((6,11)\).

Find the value of \(k\).   (3 marks)

Show Answers Only

\(k=6\)

Show Worked Solution

\(y=(x-1)(x-5)\)

\(\text{Translate \(k\) units to the right:}\)

\(y \rightarrow y^{\prime}=(x-k-1)(x-k-5)\)

\(\text{Translate \(k\) units vertically up:}\)

\(y^{\prime} \rightarrow y^{\prime \prime}=(x-k-1)(x-k-5)+k\)

\(y^{\prime \prime} \ \text{passes through}\ (6,11):\)

\(11=(6-k-1)(6-k-5)+k\)

\(11=(5-k)(1-k)+k\)

\(11=5-6 k+k^2+k\)

\(0=k^2-5 k-6\)

\(0=(k-6)(k+1)\)

\(\therefore k=6 \quad(k>0)\)

Trigonometry, 2ADV T1 2025 HSC 28

A farmer wants to use a straight fence to divide a circular paddock of radius 10 metres into two segments. The smaller segment is \(\dfrac{1}{4}\) of the paddock and is shaded in the diagram. The fence subtends an angle of \(\theta\) radians at the centre of the circle as shown.
 

  1. Show that  \(\theta=\sin \theta+\dfrac{\pi}{2}\).   (2 marks)

  2. The graph of  \(y=\sin \theta+\dfrac{\pi}{2}\) is shown.
     
  4. Use the graph and the result in part (a) to estimate the arc length of the smaller segment to the nearest metre.   (2 marks)

Show Answers Only

a.   \(\text{See Worked Solutions}\)

b.   \(\text{Arc length} \ \approx 23 \ \text{metres.}\)

Show Worked Solution

a.    \(\text{Area of paddock} =\pi \times 10^2=100 \pi\)

\(\text{Area of segment} =\dfrac{1}{4} \times 100 \pi =25 \pi\)

\(\text{Area of sector} =\dfrac{\theta}{2 \pi} \times 100 \pi=50 \theta\)

\(\text{Area of triangle} =\dfrac{1}{2} ab \, \sin C=50 \, \sin \theta\)

\(\text{Equating sector areas:}\)

\(25 \pi\) \(=50 \theta-50\, \sin \theta\)
\(50 \theta\) \(=25 \pi+50 \, \sin \theta\)
\(\theta\) \(=\dfrac{\pi}{2}+\sin \theta\)

 
b.    \(\text{Find where} \ \ \theta=\sin \theta+\dfrac{\pi}{2}\)

\(\text {Intersection occurs where: }\)

\(y=\theta \ \ \text{intersects with} \ \ y=\sin \theta+\dfrac{\pi}{2}\)

\(\text{At intersection (from graph):} \ \ \theta \approx 2.3 \ \text{radians}\)

\(\text{Arc length} \ \approx 10 \times 2.3 \approx 23 \ \text{metres.}\)

Calculus, 2ADV C4 2025 HSC 27

The shaded region is bounded by the graph  \(y=\left(\dfrac{1}{2}\right)^x\), the coordinate axes and  \(x=2\).
 

  1. Use two applications of the trapezoidal rule to estimate the area of the shaded region.   (2 marks)

  2. Show that the exact area of the shaded region is  \(\dfrac{3}{4 \ln 2}\).   (2 marks)

  3. Using your answers from part (a) and part (b), deduce  \(e<2 \sqrt{2}\).   (2 marks)

Show Answers Only

a.   \(A\approx \dfrac{9}{8}\ \text{units}^2\)
 

b.     \(\text{Area}\) \(=\displaystyle \int_0^2\left(\frac{1}{2}\right)^x d x\)
    \(=\left[-\dfrac{2^{-x}}{\ln 2}\right]_0^2\)
    \(=-\dfrac{2^{-2}}{\ln 2}+\dfrac{1}{\ln 2}\)
    \(=-\dfrac{1}{4 \ln 2}+\dfrac{1}{\ln 2}\)
    \(=\dfrac{3}{4 \ln 2}\)

 

c.    \(\text{Trapezoidal estimate assumes a straight line (creating a trapezium)}\)

\(\text{between (0,1) and \((2,\dfrac{1}{4})\)}\)

\(\Rightarrow \ \text{Area using trap rule > Actual area}\)

\(\dfrac{9}{8}\) \(>\dfrac{3}{4 \ln 2}\)  
\(36\, \ln 2\) \(>24\)  
\(\ln 2\) \(>\dfrac{2}{3}\)  
\(e^{\small \dfrac{2}{3}}\) \(>2\)  
\(e\) \(>2^{\small \dfrac{3}{2}}\)  
\(e\) \(>2 \sqrt{2}\)  
Show Worked Solution

a.  

\begin{array}{|c|c|c|c|}
\hline \ \ x \ \  & \ \ 0 \ \  & \ \ 1 \ \  & \ \ 2 \ \  \\
\hline y & 1 & \dfrac{1}{2} & \dfrac{1}{4} \\
\hline
\end{array}

\(A\) \(\approx \dfrac{h}{2}\left[1 \times 1+2 \times \dfrac{1}{2}+1 \times \dfrac{1}{4}\right]\)
  \(\approx \dfrac{1}{2}\left(\dfrac{9}{4}\right)\)
  \(\approx \dfrac{9}{8}\ \text{units}^2\)
 
b.     \(\text{Area}\) \(=\displaystyle \int_0^2\left(\frac{1}{2}\right)^x d x\)
    \(=\left[-\dfrac{2^{-x}}{\ln 2}\right]_0^2\)
    \(=-\dfrac{2^{-2}}{\ln 2}+\dfrac{1}{\ln 2}\)
    \(=-\dfrac{1}{4 \ln 2}+\dfrac{1}{\ln 2}\)
    \(=\dfrac{3}{4 \ln 2}\)

  

c.    \(\text{Trapezoidal estimate assumes a straight line (creating a trapezium)}\)

\(\text{between (0,1) and \((2,\dfrac{1}{4})\)}\)

\(\Rightarrow \ \text{Area using trap rule > Actual area}\)

\(\dfrac{9}{8}\) \(>\dfrac{3}{4 \ln 2}\)  
\(36\, \ln 2\) \(>24\)  
\(\ln 2\) \(>\dfrac{2}{3}\)  
\(e^{\small \dfrac{2}{3}}\) \(>2\)  
\(e\) \(>2^{\small \dfrac{3}{2}}\)  
\(e\) \(>2 \sqrt{2}\)  

Calculus, 2ADV C4 2025 HSC 25

  1. Show that  \(\dfrac{d}{d x}(\sin x-x\, \cos x)=x\, \sin x\).   (2 marks)
  2. Hence, find the value of  \(\displaystyle\int_0^{2025 \pi} x\, \sin x \, dx\).   (2 marks)

  3. The regions bounded by the \(x\)-axis and the graph of  \(y=x\, \sin x\)  for  \(x \geq 0\)  are shown.
     

  1. Let  \(A_n=\displaystyle \int_{(n-1) \pi}^{n \pi} x\, \sin x \,dx\),  where \(n\) is a positive integer.
  2. It can be shown that  \(\left|A_n\right|=(2 n-1) \pi\).  (Do NOT prove this.)
  3. Find the exact total area of the regions bounded by the curve  \(y=x \sin x\), and the \(x\)-axis between  \(x=0\)  and  \(x=2025 \pi\).   (2 marks)  
Show Answers Only

a.   \(\text{See Worked Solutions}\)

b.   \(2025 \pi\)

c.   \(4\,100\,625 \pi \ \text{units}^2\)

Show Worked Solution
a.     \(\dfrac{d}{dx}(\sin x-x\, \cos x)\) \(=\dfrac{d}{dx} \sin x-\dfrac{d}{dx} x\, \cos x\)
    \(=\cos x+x\, \sin x-\cos x\)
    \(=x\, \sin x\)

 

b.     \(\displaystyle\int_0^{2025 \pi} x\, \sin x\) \(=\Big[\sin x-x\, \cos x\Big]_0^{2025 \pi}\)
    \(=\Big[(\sin (2025\pi)-2025 \pi \times \cos (2025 \pi))-0\Big]\)
    \(=0-2025 \pi \times -1\)
    \(=2025 \pi\)

 

c.    \(\text{Area}=\displaystyle \int_0^\pi x\, \sin x \, dx+\left|\int_\pi^{2 \pi} x\, \sin x \, dx\right|+\cdots+\int_{2024 \pi}^{2025 \pi} x\, \sin x \, dx\)

\(\text{Using}\ \ \left|A_n\right|=(2n-1) \pi:\)

\(A_1=\pi, A_2=3 \pi, A_3=5 \pi, \ldots, A_{2025}=4049 \pi\)

\begin{aligned}
\rule{0pt}{2.5ex} \text{Area}& =\underbrace{\pi+3 \pi+5 \pi+\ldots+4049 \pi}_{\text {AP where } a=\pi, \ l=4049 \pi, \ n=2025} \\
\rule{0pt}{4.5ex} & =\frac{2025}{2}(\pi+4049 \pi) \\
\rule{0pt}{3.5ex} & =4\,100\,625 \pi \ \text{units }^2
\end{aligned}

Statistics, 2ADV S3 2025 HSC 23

  1. In a flock of 12 600 sheep, the ratio of males to females is \(1:20\).
  2. The weights of the male sheep are normally distributed with a mean of 76.2 kg and a standard deviation of 6.8 kg.
  3. In the flock, 15 of the male sheep each weigh more than \(x\) kg. 
  4. Find the value of \(x\).   (4 marks)

  5. The weights of the female sheep are also normally distributed but have a smaller mean and smaller standard deviation than the weights of male sheeр.
  6. Explain whether it could be expected that 300 of the females from the flock each weigh more than \(x\) kg, where \(x\) is the value found in part (a).   (1 mark)

Show Answers Only

a.   \(x=89.8 \ \text{kg}\)

b.    \(\text{Female sheep} \ \%=\dfrac{300}{12\,000}=0.025 \%\ (z \text{-score = 2)}\)

\(\text{Given} \ \ \bar{x}_f<\bar{x}_m \ \ \text{and} \ \ s_f<s_m\)

\(\text{Consider the value of}\ x_f\ \text{when \(z\)-score = 2}:\)

\(\Rightarrow x_f=\bar{x}_f+2 \times s_f \leq x_m\)

\(\therefore \ \text{It is not expected that  300 females weigh > 89.8 kg.}\)

Show Worked Solution

a.    \(12\,600 \ \text{sheep} \ \Rightarrow \ \text{male : female}=1:20\)

\(\text{Number of sheet in “1 part”} = \dfrac{12\,600}{21}=600\)

\(\Rightarrow \ \text{male : female}=600:12\,000\)

\(\text{male sheep} \ \%=\dfrac{15}{600}=0.025\%\)

\(z \text{-score }(0.025 \%)=2\)

\(\text{Using } \ z=\dfrac{x-\bar{x}}{s}, \ \text{find} \ \ x_m:\)

\(2\) \(=\dfrac{x_m-76.2}{6.8}\)  
\(x_m\) \(=76.2+2 \times 6.8=89.8 \ \text{kg}\)  

 
b.    \(\text{Female sheep} \ \%=\dfrac{300}{12\,000}=0.025 \%\ (z \text{-score = 2)}\)

\(\text{Given} \ \ \bar{x}_f<\bar{x}_m \ \ \text{and} \ \ s_f<s_m\)

\(\text{Consider the value of}\ x_f\ \text{when \(z\)-score = 2}:\)

\(\Rightarrow x_f=\bar{x}_f+2 \times s_f \leq x_m\)

\(\therefore \ \text{It is not expected that  300 females weigh > 89.8 kg.}\)

Trigonometry, 2ADV T2 2025 HSC 22

Prove that

\(\dfrac{\sin ^4 \theta+\cos ^4 \theta}{\sin ^2 \theta\, \cos ^2 \theta}+2=\sec ^2 \theta\, \operatorname{cosec}^2 \theta\).   (2 marks)

Show Answers Only

\(\text{See Worked Solutions}\)

Show Worked Solution

\(\text{Note: RHS } =\sec ^2 \theta\, \operatorname{cosec}^2 \theta=\dfrac{1}{\sin ^2 \theta\, \cos ^2 \theta}\)

\(\text{RHS}\) \(=\dfrac{\sin ^4 \theta+\cos ^4 \theta}{\sin ^2 \theta \cos ^2 \theta}+2\)
  \(=\dfrac{\sin ^4 \theta+\cos ^4 \theta+2 \sin ^2 \theta \cos ^2 \theta}{\sin ^2 \theta\, \cos ^2 \theta}\)
  \(=\dfrac{\left(\sin ^2 \theta+\cos ^2 \theta\right)^2}{\sin ^2 \theta\, \cos ^2 \theta}\)
  \(=\dfrac{1}{\sin ^2 \theta\, \cos ^2 \theta}\)
  \(=\sec ^2 \theta\, \operatorname{cosec} ^2 \theta\)

Financial Maths, 2ADV M1 2025 HSC 20

The table shows future value interest factors for an annuity of $1.

Lin invests a lump sum of $21 000 for 7 years at an interest rate of 6% per annum, compounding monthly.

Yemi wants to achieve the same future value as Lin by using an annuity. Yemi plans to deposit a fixed amount into an investment account at the end of each month for 7 years. The investment account pays 6% per annum, compounding monthly.

Using the table provided, determine how much Yemi needs to deposit each month.   (3 marks)

Show Answers Only

\($306.78\)

Show Worked Solution

\(r=\dfrac{0.06}{12}=0.005, \ n=12 \times 7=84\)

\(\text{Lin’s investment:}\)

\(F V=21\,000(1+0.005)^{84}=31\,927.76\)
 

\(\text{Yemi’s investment:}\)

\(\text{Annuity factor:} \ 104.07393\)

\(\text{Annuity} \times 104.07393\) \(=$31\,927.76\)
\(\text{Annuity}\) \(=\dfrac{31\,927.76}{104.07393}=$306.78\)

Functions, 2ADV F1 2025 HSC 18

Find the range of \(g(f(x))\),  given  \(f(x)=\dfrac{3}{x-1}\)  and  \(g(x)=x+5\).   (2 marks)

Show Answers Only

\(\text{Range} \ g(f(x)): \text{All} \  y, \  y \neq 5 \ \ \text{or} \ \ \{-\infty<y<5\} \cup\{5<y<\infty\}.\)

Show Worked Solution

\(f(x)=\dfrac{3}{x-1}, \ \ g(x)=x+5\)

\(g(f(x))=\dfrac{3}{x-1}+5 \Rightarrow \ \text{vertical translation +5 of} \ f(x)\)

\(\text{Range} \ f(x):\ \text{All} \ y, \ y \neq 0\)

\(\text{Range} \ g(f(x)): \text{All} \  y, \  y \neq 5 \ \ \text{or} \ \ \{-\infty<y<5\} \cup\{5<y<\infty\}.\)

Financial Maths, 2ADV M1 2025 HSC 17

A borrower obtains a reducing-balance loan of $800 000 to buy a house.

Interest is charged at 0.5% monthly, compounded monthly.

On the last day of each month, interest is added to the balance owing on the loan and then the monthly repayment of $5740 is made.

Let \(\$ A_n\) be the balance owing on the loan at the end of \(n\) months.

  1. Show that  \(A_2=800\,000(1.005)^2-5740(1.005)-5740\).   (2 marks)

  2. Show that  \(A_n=1\,148\,000-348\,000(1.005)^n\).   (3 marks)

  3. After how many months will the balance owing on the loan first be less than $400 000?   (2 marks)

Show Answers Only

a.   \(\text{See Worked Solutions.}\)

b.   \(\text{See Worked Solutions.}\)

c.   \(\text{154 months}\)

Show Worked Solution
a.     \(A_1\) \(=800\,000(1.005)-5740\)
  \(A_2\) \(=A_1 \times (1.005)-5470\)
    \(=[800\,000(1.005)-5740](1.005)-5740\)
    \(=800\,000(1.005)^2-5740(1.005)-5470\)

 

b.     \(A_3\) \(=800\,000(1.005)^3-5740(1.005)^2-5740(1.005)-5740\)
  \(A_n\) \(=800\,000(1.005)^n-5740(1.005)^{n-1} \ldots -5740(1.005)-5740\)
    \(=800\,000(1.005)^n-5740 \underbrace{\left(1.005+1.005^2+\cdots+1.005^{n-1}\right)}_{\text {GP where} \  a=1 ,  r=1.005}\)
    \(=800\,000(1.005)^n-5740\left(\dfrac{a\left(r^n-1\right)}{r-1}\right)\)
    \(=800\,000(1.005)^n-5740\left(\dfrac{1.005^n-1}{0.005}\right)\)
    \(=800\,000(1.005)^n-1\,148\,000\left(1.005^n-1\right)\)
    \(=1\,148\,000-1\,148\,000(1.005)^n+800\,000(1.005)^n\)
    \(=1\,148\,000-348\,000(1.005)^n\)

 

c.    \(\text{Find} \  n \ \text{such that} \ A_n < 400\,000:\)

\(1\,148\,000-348\,000(1.005)^n\) \(< 400\,000\)
\(348\,000(1.005)^n\) \(>1\,148\,000-400\,000\)
\(1.005^n\) \(>\dfrac{748\,000}{348\,000}\)
\(n \times \ln 1.005\) \(>\ln \left(\frac{187}{87}\right)\)
\(n\) \(>\dfrac{\ln \left(\frac{187}{87}\right)}{\ln 1.005}\)
\(n\) \(>153.42\)

 
\(\therefore \ \text{Balance owing is less than \$400 000 after 154 months.}\)

Calculus, 2ADV C3 2025 HSC 16

Consider the function  \(f(x)=\dfrac{x^2}{e^x}\).

  1. Find the stationary points of the function and determine their nature.   (4 marks)

  2. A partially completed graph of  \(f(x)=\dfrac{x^2}{e^x}\)  is shown.
  3. Use your answer from part (a) to complete the graph.   (1 mark)


     

Show Answers Only

a.   \(\text{MIN at}\ \ (0,0)\)

\(\text{MAX at}\ \ \left(2,\dfrac{4}{e^2}\right)\)

b.   
     

Show Worked Solution
a.     \(f(x)\) \(=\dfrac{x^2}{e^x}\)
  \(f^{\prime}(x)\) \(=2 x \cdot e^x-e^x \cdot x^2\)
    \(=\dfrac{x e^x(2-x)}{e^{2 x}}\)
    \(=\dfrac{x(2-x)}{e^x}\)

 
\(\text{Find} \ x\ \text{when} \ \ f^{\prime}(x)=0:\)

\(x(2-x)=0\)

\(x=0 \ \text {or} \ 2\)

\(\text{When} \ \ x=0 \ \Rightarrow \ f(0)=0\)

\(\text {When}\ \  x=2 \ \Rightarrow \ f(2)=\dfrac{4}{e^2}\)

\(\text {Checking nature of SP’s:}\)

\begin{array}{|c|c|c|c|c|c|}
\hline x & -1 & 0 & 1 & 2 & 3 \\
\hline f^{\prime}(x) & -3 e & 0 & \dfrac{1}{e} & 0 & -\dfrac{3}{e^3} \\
\hline
\end{array}

\(\therefore \ \text{MIN at}\ \ (0,0)\)

\(\quad \ \ \text{MAX at}\ \ \left(2,\dfrac{4}{e^2}\right)\)
 

b.
     

Statistics, STD2 S4 2025 25

In a research study, participants were asked to record the number of minutes they spent watching television and the number of minutes they spent exercising each day over a period of 3 months. The averages for each participant were recorded and graphed.

 

  1. Describe the bivariate dataset in terms of its form and direction.   (2 marks)
  2. Form:  ..................................................................
  3. Direction:  ............................................................

The equation of the least-squares regression line for this dataset is

\(y=64.3-0.7 x\)

  1. Interpret the values of the slope and \(y\)-intercept of the regression line in the context of this dataset.   (2 marks)

  2. Jo spends an average of 42 minutes per day watching television.
  3. Use the equation of the regression line to determine how many minutes on average Jo is expected to exercise each day.   (1 mark)

  4. Explain why it is NOT appropriate to extrapolate the regression line to predict the average number of minutes of exercise per day for someone who watches an average of 2 hours of television per day.   (1 mark)

Show Answers Only

a.    \(\text{Form: Linear. Direction: Negative}\)

Show Worked Solution

a.    \(\text{Form: Linear}\)

\(\text{Direction: Negative}\)
 

b.    \(\text{Slope}=-0.7\)

\(\text{This means that for each added minute of watching television per day, a participant, on average,}\)

\(\text{will exercise for 0.7 minutes less.}\)

\(y \text{-intercept}=64.3\)

\(\text{If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.}\)
 

c.    \(\text{At} \ \ x=42:\)

\(y=64.3-0.7 \times 42=34.9\)

\(\therefore \ \text{Jo is expected to exercise for 34.9 minutes}\)
 

d.    \(\text{At} \ \  x=120\text{(2 hours),} \ \ y=64.3-0.7 \times 120=-19.7\)

\(\text{The model predicts a negative value for time spent exercising, which is not possible.}\)

Statistics, 2ADV S2 2025 HSC 14

In a research study, participants were asked to record the number of minutes they spent watching television and the number of minutes they spent exercising each day over a period of 3 months. The averages for each participant were recorded and graphed.
 

 

  1. Describe the bivariate dataset in terms of its form and direction.   (2 marks)
  2. Form:  ..................................................................
  3. Direction:  ............................................................

The equation of the least-squares regression line for this dataset is

\(y=64.3-0.7 x\)

  1. Interpret the values of the slope and \(y\)-intercept of the regression line in the context of this dataset.   (2 marks)

  2. Jo spends an average of 42 minutes per day watching television.
  3. Use the equation of the regression line to determine how many minutes on average Jo is expected to exercise each day.   (1 mark)

  4. Explain why it is NOT appropriate to extrapolate the regression line to predict the average number of minutes of exercise per day for someone who watches an average of 2 hours of television per day.   (1 mark)

Show Answers Only

a.    \(\text{Form: Linear}\)

\(\text{Direction: Negative}\)
 

b.    \(\text{Slope}=-0.7\)

\(\text{This means that for each added minute of watching television per day, a participant, on average,}\)

\(\text{will exercise for 0.7 minutes less.}\)

\(y \text{-intercept}=64.3\)

\(\text{If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.}\)
 

c.    \(\text{34.9 minutes}\)
 

d.    \(\text{At} \ \  x=120\ \text{(2 hours),} \ \ y=64.3-0.7 \times 120=-19.7\)

\(\text{The model predicts a negative value for time spent exercising, which is not possible.}\)

Show Worked Solution

a.    \(\text{Form: Linear}\)

\(\text{Direction: Negative}\)
 

b.    \(\text{Slope}=-0.7\)

\(\text{This means that for each added minute of watching television per day, a participant, on average,}\)

\(\text{will exercise for 0.7 minutes less.}\)

\(y \text{-intercept}=64.3\)

\(\text{If someone watches no television, the LSRL predicts they will exercise for 64.3 minutes per day.}\)
 

c.    \(\text{At} \ \ x=42:\)

\(y=64.3-0.7 \times 42=34.9\)

\(\therefore \ \text{Jo is expected to exercise for 34.9 minutes}\)
 

d.    \(\text{At} \ \  x=120\ \text{(2 hours),} \ \ y=64.3-0.7 \times 120=-19.7\)

\(\text{The model predicts a negative value for time spent exercising, which is not possible.}\)

Algebra, STD2 A4 2025 HSC 20

The graph of a quadratic function represented by the equation  \(h=t^2-8 t+12\)  is shown.
 

  1. Find the values of \(t\) and \(h\) at the turning point of the graph.   (2 marks)

  2. The graph shows  \(h=12\)  when  \(t=0\).
  3. What is the other value of \(t\) for which  \(h=12\)?   (1 mark)

Show Answers Only

a.   \(\text{Turning point at} \ \ (4,-4)\)

b.   \(t=8\)

Show Worked Solution

a.    \(\text{Axis of quadratic occurs when}\ \ t= \dfrac{2+6}{2} = 4\)

\(\text{At} \ \ t=4:\)

\(h=4^2-8 \times 4+12=-4\)

\(\therefore \ \text{Turning point at} \ \ (4,-4)\)
 

b.    \(\text {When} \ \ h=12:\)

\(t^2-8 t+12\) \(=12\)
\(t(t-8)\) \(=0\)

 
\(\therefore \ \text{Other value:} \ \ t=8\)

Statistics, 2ADV S3 2025 HSC 8 MC

The minimum daily temperature, in degrees, of a town each year follows a normal distribution with its mean equal to its standard deviation. The minimum daily temperature was recorded over one year.

What percentage of the recorded minimum daily temperatures was above zero degrees?

  1. 16%
  2. 50%
  3. 68%
  4. 84%
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Consider a possible example:}\)

\(\text{Let mean min daily temperature = 8°C}\)

\(\text{Std dev = 8°C}\)

\(z\text{-score (0°C)}\ =-1\)

\(\text{Percentage above 0°C} = 50+34=84\%\)

\(\Rightarrow D\)

Probability, 2ADV S1 2025 HSC 7 MC

A ten-sided die has faces numbered 1 to 10 .

The die is constructed so that the probability of obtaining the number 1 is greater than the probability of obtaining any of the other numbers. The numbers 2 to 10 are equally likely to occur.

When the die is rolled 153 times, a 1 is obtained 72 times.

By using the relative frequency of rolling a 1, which of the following is the best estimate for the probability of rolling a 10 ?

  1. \(\dfrac{1}{17}\)
  2. \(\dfrac{1}{11}\)
  3. \(\dfrac{1}{10}\)
  4. \(\dfrac{1}{9}\)
Show Answers Only

\(A\)

Show Worked Solution

\(P(1) = \dfrac{72}{153}=\dfrac{8}{17} \)

\(\text{Let}\ \ p=P(2)=P(3) = … =P(10) \)

\(\dfrac{8}{17}+9p\) \(=1\)  
\(9p\) \(=1-\dfrac{8}{17}\)  
\(p\) \(=\dfrac{1}{17}\)  

 
\(\Rightarrow A\)

Calculus, 2ADV C1 EO-Bank 2

  1.  Find the equations of the tangents to the curve  `y = x^2-5x+6`  at the points where the curve cuts the `x`-axis.  (2 marks)

  2.  Where do the tangents intersect?  (2 marks)

Show Answers Only
  1. `y = −x+2`
    `y = x-3`
  2. `(5/2, −1/2)`
Show Worked Solution
a.   `y` `= x^2-5x+6`
  `= (x-2)(x-3)`

 
`text(Cuts)\ xtext(-axis at)\ \ x = 2\ \ text(or)\ \ x = 3`
 

`(dy)/(dx) = 2x-5`

 
`text(At)\ \ x = 2 \ => \ (dy)/(dx) = -1`

`T_1\ text(has)\ \ m = −1,\ text{through (2, 0)}`

`y -0` `= -1(x-2)`
`y` `= -x+2`

  

`text(At)\ \ x = 3 \ => \ (dy)/(dx) = 1`

`T_2\ text(has)\ \ m = 3,\ text{through (3, 0)}`

`y -0` `= 1(x-3)`
`y` `= x -3`

 

b.   `text(Intersection occurs when:)`

`-x+2` `= x-3`
`2x` `= 5`
`x` `= 5/2`

  

`y = 5/2 – 3 = −1/2`

`:.\ text(Intersection at)\ \ (5/2, −1/2)`

Calculus, 2ADV C1 EO-Bank 1

Find the equation of the tangent to the curve  \(y=e^{x^2+3x}\)  at the point where \(x=1\).  (2 marks)

Show Answers Only

`y = 5e^4x-4e^4`

Show Worked Solution
\(y\) \(=e^{x^2+3x}\)
`(dy)/(dx)` \(=(2x+3)e^{x^2+3x}\)

 
`text(When)\ x = 1,\ \ (dy)/(dx) = 5e^4`

`text(Equation of tangent through)\ (1, e^4)`

`y-e^4` `= 5e^4(x – 1)`
`y` `= 5e^4x-4e^4`

Calculus, 2ADV C1 EO-Bank 14 v1

Evaluate `f^{′}(1)`, where `f(x) = x^2 / sqrt(2x + 3)`. (4 marks)

Show Answers Only

`9 / (5sqrt5)`

Show Worked Solution

`f(x) = x^2(2x + 3)^(-1/2)`

`f^{′}(x)` `= 2x(2x + 3)^(-1/2) + x^2(-1/2)(2x + 3)^(-3/2)(2)`

`= (2x)/(sqrt(2x + 3)) – (x^2)/(2x + 3)^(3/2)`

`= [2x(2x + 3) – x^2] / (2x + 3)^(3/2)`

`= (3x^2 + 6x) / (2x + 3)^(3/2)`

`f^{′}(1)` `= (3(1)^2 + 6(1)) / (2(1) + 3)^(3/2)`

`= 9 / (5sqrt5)`

Calculus, 2ADV C1 EO-Bank 6

Use the definition of the derivative, `f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}`  to find  `f^{\prime}(x)`  if  `f(x)=x-3x^2`.   (2 marks)

Show Answers Only

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

`f^{′}(x)` `= \lim_{h->0} \frac{(x+h)-3(x+h)^2-(x-3x^2)}{h}`  
  `= \lim_{h->0} \frac{x+h-3x^2-6hx-3h^2-x+3x^2}{h}`  
  `= \lim_{h->0} \frac{h-6hx-3h^2}{h}`  
  `= \lim_{h->0} \frac{h(1-6x-3h)}{h}`  
  `= \lim_{h->0} 1-6x-3h`  
  `=1-6x`  

Show Worked Solution

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

`f^{′}(x)` `= \lim_{h->0} \frac{(x+h)-3(x+h)^2-(x-3x^2)}{h}`  
  `= \lim_{h->0} \frac{x+h-3x^2-6hx-3h^2-x+3x^2}{h}`  
  `= \lim_{h->0} \frac{h-6hx-3h^2}{h}`  
  `= \lim_{h->0} \frac{h(1-6x-3h)}{h}`  
  `= \lim_{h->0} 1-6x-3h`  
  `=1-6x`  

Calculus, 2ADV C1 2013 HSC 11b v1

Evaluate  `lim_(x->1) ((x-1)(x+2)^2)/(x^2+x-2)`.   (2 marks)

Show Answers Only

 `3`

Show Worked Solution

`lim_(x ->1) ((x-1)(x+2)^2)/(x^2+x-2)`

COMMENT: This question has been simplified as students no longer need to factorise the difference between 2 cubes (`x^3-2^3`).

`=lim_(x->1) ( (x -1)(x+2)^2)/( (x-1)(x+2)`

`=lim_(x->1) (x+2)`

`=3`

Calculus, 2ADV C4 EO-Bank 4 MC SJ

 Let  `f^(')(x)=(2)/(sqrt(2x-3))`. 

If  `f(6)=4`, then
 

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

`=>C`

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

 
`text(When)\ \ x=6, \ f(x)=4:`

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

`:. f(x) = 2sqrt(2x-3)-2`

`=>C`

Calculus, 2ADV C1 EO-Bank 1 MC v1

The derivative of  \((n^2-1) x^{3n-2}\)  can be expressed as

  1. \(3(n-1)(n^2-1) x^{3n-2}\)
  2. \(3(n-1)(n^2-1) x^{3(n-1)}\)
  3. \((3n-2) (n^2-1) x^{3(n-1)}\)
  4. \((3n-2) (n^2-1) x^{3n-2}\)
Show Answers Only

\(C\)

Show Worked Solution
\(y\) \(=(n^2-1) x^{3n-2}\)  
\(y^{′}\) \(=(3n-2) (n^2-1) x^{3n-2-1)}\)  
  \(=(3n-2) (n^2-1) x^{3(n-1)}\)  

 
\(\Rightarrow C\)

BIOLOGY, M7 2021 VCE 11

Two students designed an experiment to investigate antibiotic resistance in Escherichia coli bacteria. They began with an E. coli culture. The following procedure was conducted in a filtered air chamber using aseptic techniques:

  • On Day 0, spread 1 mL of E. coli culture onto a nutrient agar plate containing \(0 \ \mu \text{g} / \text{mL}\) (micrograms per millilitre) of the antibiotic ampicillin. Spread \(1 \ \text{mL}\) of the \(E. coli \) culture onto a separate nutrient agar plate containing \(1 \ \mu \text{g} / \mathrm{mL}\) of ampicillin. Cover each plate with an airtight lid.
  • On Day 1, transfer a sample of bacteria from one of the Day 0 plates to one of the Day 1 plates containing \(2 \ \mu \text{g} / \text{mL}\) of ampicillin. Transfer a sample of bacteria from the other Day 0 plate to the other Day 1 plate, which also contains \(2 \ \mu \text{g} / \text{mL}\) of ampicillin. Cover as before.
  • On Day 2, transfer a sample of bacteria from one of the Day 1 plates to one of the Day 2 plates containing \(4 \ \mu \text{g} / \text{mL}\) of ampicillin. Transfer a sample of bacteria from the other Day 1 plate to the other Day 2 plate, which also contains \(4 \ \mu \text{g} / \text{mL}\) of ampicillin. Cover as before.
  • On Day 3, transfer a sample of bacteria from one of the Day 2 plates to one of the Day 3 plates containing \(6 \ \mu \text{g} / \text{mL}\) of ampicillin. Transfer a sample of bacteria from the other Day 2 plate to the other Day 3 plate, which also contains \(6 \ \mu \text{g} / \text{mL}\) of ampicillin. Cover and seal the plates.

All plates were incubated at 37 °C for each 24-hour period. Used plates were refrigerated until the end of the experiment. They were then photographed to compare the amount of bacterial growth and disposed of safely.

The students drew a diagram (Figure 1) to help explain the experimental design and to show their predicted results in each condition at the end of each day.
 

  1. Identify any two controlled variables for this experiment.   (2 marks)
  1. Write a suitable hypothesis for this experiment.   (2 marks)
The refrigerated plates kept from Days 0,1,2 and 3 of the experiment were photographed. The diagrams in Figure 2 represent the bacterial growth seen in the photographs.
   

 

  1.  i. Analyse the results of the experiment shown in Figure 2.   (3 marks)
  2. ii. Explain whether the results of the students' experiment shown in Figure 2 support the predicted results shown in Figure 1.   (2 marks)
Show Answers Only

a.    Examples of suitable responses included two of the following:

  • volume of E. coli first applied to plates
  • type of nutrient agar used
  • size of agar plates used
  • batch or source of ampicillin
  • duration of incubation
  • temperature of incubation
  • exposure time to each concentration of ampicillin.

b.    Hypothesis:

  • If E. coli is exposed to low OR 1 µg/ml concentration of ampicillin, it will exhibit increased growth when exposed to higher levels of ampicillin.

c.i.  Experiment analysis:

  • No ampicillin: More E. coli growth on Day 0.
  • 2 µg/mL ampicillin: No effect on bacterial growth.
  • Days 2/3: Fewer colonies for both groups.
  • Higher ampicillin (4 µg/mL, 6 µg/mL): Significant effect, killed many E. coli.
  • More growth than predicted on all plates.
  • By end: More ampicillin-resistant bacteria on experimental plate.
  • 1 µg/mL ampicillin: Initial exposure led to greater number of resistant E. coli.

c.ii.  If the experimental results did not support the predicted results:

  • E. coli was still present on both plates on Day 3.
  • This shows exposing E. coli to 1 µg/ml ampicillin does not increase its resistance to ampicillin.
  • Resistant E. coli grew regardless of initial exposure to ampicillin.

Any two of the following, if the experimental results did support the predicted results:

  • The number of colonies in both groups decreased over time.
  • More E. coli developed ampicillin resistance in the experimental group.
  • There were more colonies in the experimental group than in the control group on Day 3.
  • There were more colonies than expected as ampicillin concentration increased.
  • No growth was expected at the end of Day 3 for the control group.
Show Worked Solution

a.    Examples of suitable responses included two of the following:

  • volume of E. coli first applied to plates
  • type of nutrient agar used
  • size of agar plates used
  • batch or source of ampicillin
  • duration of incubation
  • temperature of incubation
  • exposure time to each concentration of ampicillin.

b.    Hypothesis:

  • If E. coli is exposed to low OR 1 µg/ml concentration of ampicillin, it will exhibit increased growth when exposed to higher levels of ampicillin.

♦♦♦ Mean mark (b) 40%.

c.i.  Experiment analysis:

  • No ampicillin: More E. coli growth on Day 0.
  • 2 µg/mL ampicillin: No effect on bacterial growth.
  • Days 2/3: Fewer colonies for both groups.
  • Higher ampicillin (4 µg/mL, 6 µg/mL): Significant effect, killed many E. coli.
  • More growth than predicted on all plates.
  • By end: More ampicillin-resistant bacteria on experimental plate.
  • 1 µg/mL ampicillin: Initial exposure led to greater number of resistant E. coli.

♦♦♦ Mean mark (c)(i) 6%.

c.ii.  If the experimental results did not support the predicted results:

  • E. coli was still present on both plates on Day 3.
  • This shows exposing E. coli to 1 µg/ml ampicillin does not increase its resistance to ampicillin.
  • Resistant E. coli grew regardless of initial exposure to ampicillin.

Any two of the following, if the experimental results did support the predicted results:

  • The number of colonies in both groups decreased over time.
  • More E. coli developed ampicillin resistance in the experimental group.
  • There were more colonies in the experimental group than in the control group on Day 3.
  • There were more colonies than expected as ampicillin concentration increased.
  • No growth was expected at the end of Day 3 for the control group.

♦♦ Mean mark (c)(ii) 50%.

CHEMISTRY, M1 EQ-Bank 10 MC

250,000 tonnes of iron ore contains 3.2% w/w of magnetite, \(\ce{Fe3O4}\).

What is the mass of magnetite?

  1. \(\text{8000 kg}\)
  2. \(8.0 \times 10^{6}\ \text{kg}\)
  3. \(8.0 \times 10^{9}\ \text{kg}\)
  4. \(8.0 \times 10^{12}\ \text{kg}\)
Show Answers Only

\(B\)

Show Worked Solution
  • Converting tonnes to kilograms:
  •     \(250\ 000\ \text{t} = 250 \ 000\ 000\ \text{kg}\)
  • Calculating the mass of magnetite:
  •     \(0.032 \times 2.5 \times 10^8\ \text{kg} = 8.0 \times 10^6\ \text{kg}\)

\(\Rightarrow B\)

CHEMISTRY, M1 EQ-Bank 6 MC

Elements \(\ce{A}\) and \(\ce{B}\) are in the same period of the Periodic Table. Element \(\ce{A}\) has 1 electron in its outer shell, and element \(\ce{B}\) has 6 electrons in its outer shell.

What is the likely formula of the compound they form together?

  1. \(\ce{A2B}\)
  2. \(\ce{AB2}\)
  3. \(\ce{A2B6}\)
  4. \(\ce{A6B2}\)
Show Answers Only

\(A\)

Show Worked Solution
  • \(\ce{A}\) (1 valence electron) → Group 1 → forms \(\ce{A^+}\).
  • \(\ce{B}\) (6 valence electrons) → Group 16 → forms \(\ce{B^2-}\) ions.
  • Hence they will form the compound with the formula \(\ce{A2B}\).

\(\Rightarrow A\)

CHEMISTRY, M1 EQ-Bank 4 MC

The formula for sodium thiosulfate is \(\ce{Na2S2O3}\).

What is the formula for aluminium thiosulfate?

  1. \(\ce{AlS2O3}\)
  2. \(\ce{Al2(S2O3)3}\)
  3. \(\ce{Al3S2O3}\)
  4. \(\ce{Al(S2O3)2}\)
Show Answers Only

\(B\)

Show Worked Solution
  • From the sodium thiosulfate ion we can identify that it contains a sodium ion \(\ce{Na^+}\) and the thiosulfate ion \(\ce{S2O3^2-}\).
  • The aluminium ion is \(\ce{Al^3+}\).
  • Hence the correct formula for aluminium thiosulfate ion is \(\ce{Al2(S2O3)3}\).

\(\Rightarrow B\)