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Trigonometry, 2ADV T3 2025 MET1 3

Let  \(f(x)=2 \cos (2 x)+1\)  over the domain \(x \in\left[0, 2 \pi \right]\).

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

  2. Solve  \(f(x)=0\)  for \(x\).   (3 marks)

  3. Sketch the graph of  \(y=f(x)\)  for  \(x \in\left[\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\right]\) on the axes below.
  4. Label the endpoints with their coordinates.   (2 marks)

Show Answers Only

a.    \(\text{Range of } f(x):-1 \leqslant y \leqslant 3\)

b.    \(x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}\)

c.   

   

Show Worked Solution

a.    \(\text{Amplitude}=2 \ \ \text{about} \ \ y=1.\)

\(\text{Range of } f(x):\ -1 \leqslant y \leqslant 3\)
 

b.     \(2 \cos (2 x)+1\) \(=0\)
  \(\cos (2 x)\) \(=-\dfrac{1}{2}\)

 
\(\text{Base angle}=\dfrac{\pi}{3}\)

\(2x=\pi-\dfrac{\pi}{3}, \pi+\dfrac{\pi}{3}, \cdots\)

\(2x=\dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{8 \pi}{3}, \dfrac{10 \pi}{3}\)

  \(x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}\)
  

c.   

   

♦ Mean mark (c) 49%.

Calculus, MET1 2025 VCAA 7

Let \(f: R \rightarrow R, f(x)=x^3-x^2-16 x-20\).

  1. Verify that  \(x=5\)  is a solution of  \(f(x)=0\).   (1 mark)

  2. Express \(f(x)\) in the form  \((x+d)^2(x-5)\), where \(d \in R\).   (2 marks)

  3. Consider the graph of  \(y=f(x)\), as shown below.
  4. Complete the coordinate pairs of all axial intercepts of  \(y=f(x)\).   (1 mark)

     

   

  1. Let \(g: R \rightarrow R, g(x)=x+2\).
    1. State the coordinates of the stationary point of inflection for the graph of  \(y=f(x) g(x)\).   (1 mark)

    2. Write down the values of \(x\) for which  \(f(x) g(x) \geq 0\).   (1 mark)

Show Answers Only

a.    \(f(5)=5^3-5^2-16 \times 5-20 =0\)

b.    \(f(x)=(x+2)^2(x-5)\)

c.   
       

d.i.   \((-2,0)\)

d.ii.  \(x \in(-\infty,-2] \cup[5, \infty)\)

Show Worked Solution

a.    \(f(x)=x^3-x^2-16 x-20\)

\(f(5)=5^3-5^2-16 \times 5-20=125-25-80-20=0\) 

\(\therefore x=5\ \ \text{is a solution of}\ f(x)\).
 

b.    \(\text{By long division:}\)
 
           

\(f(x)=\left(x^2+4 x+4\right)(x-5)=(x+2)^2(x-5)\)
 

c.   
       
 

d.i.    \(y\) \(=f(x) \cdot g(x)\)
    \(=(x+2)^2(x-5)(x+2)\)
    \(=(x+2)^3(x-5)\)

 

\((x+2)^3\ \text{factor}\ \Rightarrow\ \text{SP of inflection at} \ (-2,0)\)

♦ Mean mark (d.i) 44%.
♦♦ Mean mark (d.ii) 27%.
 

d.ii.
     

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

\(f(x) g(x) \geqslant 0 \ \ \text{when}\ \  x \leqslant-2\ \cup\  x \geqslant 5\)

\(x \in(-\infty,-2] \cup[5, \infty) \ \text{also correct.}\)

Probability, MET1 2025 VCAA 4

The probability distribution for the discrete random variable \(X\) is given in the table below, where \(k\) is a positive real number.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}x \rule[-1ex]{0pt}{0pt}& \quad \quad 0 \quad \quad & \quad \quad 1 \quad \quad & \quad \quad 2 \quad \quad & \quad \quad 3 \quad \quad \\
\hline
\rule{0pt}{2.5ex}\operatorname{Pr}(X=x) \rule[-1ex]{0pt}{0pt}& \dfrac{4}{k} &\dfrac{2 k}{75} &\dfrac{k}{75} & \dfrac{2}{k} \\
\hline
\end{array}

  1. Show that  \(k=10\)  or  \(k=15\).   (2 marks)

  2. Let  \(k=15\).
    1. Find \(\operatorname{Pr}(X>1)\).   (1 mark)

    2. Find \(E (X)\).   (1 mark)

Show Answers Only

a.    \(\text{Sum of probabilities\(=1\):}\)

\(\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}\) \(=1\)
\(\dfrac{6}{k}+\dfrac{3 k}{75}\) \(=1\)
\(3 k^2+450\) \(=75k\)
\(3 k^2-75 k+450\) \(=0\)
\(k^2-25 k+150\) \(=0\)
\((k-10)(k-15)\) \(=0\)

  
\(\therefore k=10 \ \ \text{or 15}\)

b.i.   \(\operatorname{Pr}(X>1)=\dfrac{1}{3}\) 

b.ii.  \(E(X)=\dfrac{90}{75}\)

Show Worked Solution

a.    \(\text{Sum of probabilities\(=1\):}\)

\(\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}\) \(=1\)
\(\dfrac{6}{k}+\dfrac{3 k}{75}\) \(=1\)
\(3 k^2+450\) \(=75k\)
\(3 k^2-75 k+450\) \(=0\)
\(k^2-25 k+150\) \(=0\)
\((k-10)(k-15)\) \(=0\)

 

\(\therefore k=10 \ \ \text{or 15}\)
 

b.i.   \(\operatorname{Pr}(X>1)\) \(=\operatorname{Pr}(X=2)+\operatorname{Pr}(X=3)\)
    \(=\dfrac{15}{75}+\dfrac{2}{15}\)
    \(=\dfrac{1}{3}\)

 

b.ii.   \(E(X)\) \(=0 \times \dfrac{4}{15}+1 \times \dfrac{30}{75}+2 \times \dfrac{15}{75}+3 \times \dfrac{2}{15}\)
    \(=\dfrac{30}{75}+\dfrac{30}{75}+\dfrac{30}{75}\)
    \(=\dfrac{90}{75}\)

BIOLOGY, M8 EQ-Bank 5 MC

Which coordination system detects changes in blood glucose and sends signals to effector organs to restore homeostasis?

  1. Circulatory
  2. Digestive
  3. Hormonal
  4. Neural
Show Answers Only

`D`

Show Worked Solution
  • D is correct: Neural pathways detect changes and send signals to effectors to restore homeostasis.

Other Options:

  • A is incorrect: Circulatory system transports substances; does not detect or signal.
  • B is incorrect: Digestive system processes nutrients; not a coordination system.
  • C is incorrect: Hormonal is a valid coordination system but uses chemical messengers via blood, not neural signals.

Graphs, MET1 2025 VCAA 3

Let  \(f:[0,2 \pi] \rightarrow R, f(x)=2 \cos (2 x)+1\).

  1. State the range of \(f\).   (1 mark)

  2. Solve  \(f(x)=0\)  for \(x\).   (3 marks)

  3. Sketch the graph of  \(y=f(x)\)  for  \(x \in\left[\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\right]\) on the axes below.
  4. Label the endpoints with their coordinates.   (2 marks)

Show Answers Only

a.    \(\text{Range of } f(x):-1 \leqslant y \leqslant 3\)

b.    \(x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}\)

c.   

   

Show Worked Solution

a.    \(\text{Amplitude}=2 \ \ \text{about} \ \ y=1.\)

\(\text{Range of } f(x):\ -1 \leqslant y \leqslant 3\)
 

b.     \(2 \cos (2 x)+1\) \(=0\)
  \(\cos (2 x)\) \(=-\dfrac{1}{2}\)

 
\(\text{Base angle}=\dfrac{\pi}{3}\)

\(2x=\pi-\dfrac{\pi}{3}, \pi+\dfrac{\pi}{3}, \cdots\)

\(2x=\dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{8 \pi}{3}, \dfrac{10 \pi}{3}\)

  \(x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}\)
  

c.   

   

♦ Mean mark (c) 49%.

Statistics, STD2 S1 EQ-Bank 23

A boxplot for the sale prices of a sample of 203 homes is shown.
 

  1. Calculate the range of the sale price data in the boxplot.   (1 mark)

  2. Calculate the upper fence for any outliers within the sale price data of the boxplot.   (2 marks)

Show Answers Only

a.    \(\text{Range}=900\,000\)

b.    \(1\,350\,000\)

Show Worked Solution

a.    \(\text{Range}=1\,300\,000-400\,000=900\,000\)
 

b.    \(IQR=900\,000-600\,000=300\,000\)

\(Q_3=900\,000\)

\(\text{Upper Fence}\) \(=Q_3+1.5 \times IQR\)
  \(=900\,000+1.5 \times 300\,000\)
  \(=1\,350\,000\)

Networks, GEN2 2025 VCAA 15

Frances lives in a housing estate.

On the graph below the vertices represent her favourite locations, and the edges represent the roads between them.
 

  1. Calculate the sum of the degrees of all the vertices in this graph.   (1 mark)
  2. Euler's formula, \(v+f=e+2\), holds for this graph.
  3. Complete the formula by writing the appropriate numbers in the boxes below.   (1 mark)

  1. Frances is at the gym. She would like to visit each of the other locations once and end at her home.
  2. What is the mathematical term used to describe this route?   (1 mark)
  3. Using edges from the original graph, construct a spanning tree below.   (1 mark)

     


Show Answers Only

a.    \(\text{Sum of degrees}=2+4+2+3+3=14\)

b.    \(5+4=7+2\)

c.    \(\text{Hamiltonian path}\)

d.    \(\text{Spanning tree (one of many possibilities):}\)
 

Show Worked Solution

a.    \(\text{Sum of degrees}=2+4+2+3+3=14\)
 

b.    \(v+f=e+2 \ \ \Rightarrow \ \ 5+4=7+2\)
 

c.    \(\text{Hamiltonian path}\)

♦ Mean mark (c) 53%.

d.    \(\text{Spanning tree (one of many possibilities):}\)
 

Data Analysis, GEN2 2025 VCAA 2

A boxplot for the sale prices of a sample of 203 homes is shown.
 

  1. Calculate the range of the sale price data in the boxplot.   (1 mark)

  2. Calculate the upper fence for the sale price data in the boxplot.   (1 mark)

Show Answers Only

a.    \(\text{Range}=900\,000\)

b.    \(1\,350\,000\)

Show Worked Solution

a.    \(\text{Range}=1\,300\,000-400\,000=900\,000\)
 

b.    \(IQR=900\,000-600\,000=300\,000\)

\(Q_3=900\,000\)

\(\text{Upper Fence}\) \(=Q_3+1.5 \times IQR\)
  \(=900\,000+1.5 \times 300\,000\)
  \(=1\,350\,000\)

Data Analysis, GEN2 2025 VCAA 1

Table 1, below, shows the prices in dollars, price, for a sample of 20 homes sold in an inner Melbourne suburb during 2017.

The type of home sold is either an apartment or a house.

Table 1

\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex}\quad \ \textit{Price(\$)}\quad \ \rule[-1ex]{0pt}{0pt}& \textit{Type} \\
\hline \rule{0pt}{2.5ex}350\,000 \rule[-1ex]{0pt}{0pt}& \ \ \text{apartment}\ \ \\
\hline \rule{0pt}{2.5ex}490\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}500\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}620\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}720\,000 \rule[-1ex]{0pt}{0pt}\rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}830\,000 & \text{apartment}\\
\hline \rule{0pt}{2.5ex}875\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}995\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}1\,100\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}1\,520\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}800\,000 \rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}840\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}920\,000 \rule[-1ex]{0pt}{0pt}& \text{house} \\
\hline \rule{0pt}{2.5ex}920\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,010\,000\rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}1\,263\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,398\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,460\,000\rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}1\,540\,000 \rule[-1ex]{0pt}{0pt}& \text{house} \\
\hline \rule{0pt}{2.5ex}1\,540\,000 \rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline
\end{array}

  1. Find the median, in dollars, of the variable price.   (1 mark)

  2. State whether the variable type is numerical, nominal or ordinal.   (1 mark)

    1. Complete the table below by finding the standard deviation, to the nearest whole number, for the sale price of apartments in the sample.   (1 mark)

    2. Table 2
      \begin{array}{|c|c|}
      \hline \rule{0pt}{2.5ex} \quad \ \ \textbf{Type of home} \quad \ \ & \quad \textbf{Standard deviation of} \quad \\
      & \rule[-1ex]{0pt}{0pt}\textbf{sale price (\$)}\\
      \hline \rule{0pt}{2.5ex}\text{house} \rule[-1ex]{0pt}{0pt}& 300\,911 \\
      \hline \rule{0pt}{2.5ex}\text{apartment} \rule[-1ex]{0pt}{0pt}& \\
      \hline
      \end{array}
    3. Using the information in Table 2, comment on the relative spread in the distribution of the sale prices of houses compared with apartments in this sample.   (1 mark)

  4. Table 3, below, shows the percentage of houses and apartments with prices in the given ranges. Some information is missing.
  5. Use the data from Table 1 to complete Table 3.
  6. Table 3 
Show Answers Only

a.    \(\text{Median}=920\,000\)

b.    \(\text{Variable is nominal}\)

c.i.  \(\text{Std deviation = \$346 466}\)

c.ii.  \(\text {Apartment sale prices have a higher spread (standard deviation) than the}\)

\(\text{spread of house sale prices.}\)

d.

\begin{array}{|c|c|}
\hline \hline \rule{0pt}{2.5ex}\ \ \ \text {House}(\%) \ \ \ \rule[-1ex]{0pt}{0pt}& \text {Apartment}(\%) \\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 30 \\
\hline \hline \rule{0pt}{2.5ex}40 \rule[-1ex]{0pt}{0pt}& 50 \\
\hline \hline \rule{0pt}{2.5ex}60 \rule[-1ex]{0pt}{0pt}& 20 \\
\hline \hline \rule{0pt}{2.5ex}100 \rule[-1ex]{0pt}{0pt}& 100 \\
\hline
\end{array}

Show Worked Solution

a.    \(\text{Median}=\dfrac{10^{\text{th}}+11^{\text{th}}}{2}=920\,000\)
 

b.    \(\text{Type is qualitative and cannot be ordered}\)

\(\Rightarrow \ \text{Variable is nominal}\)
 

c.i.  \(\text{By calculator,}\)

\(\text{Std deviation = \$346 466}\)
 

c.ii.  \(\text {Apartment sale prices have a higher spread (standard deviation) than the}\)

\(\text{spread of house sale prices.}\)

♦ Mean mark (c.ii) 47%.

d.

\begin{array}{|c|c|}
\hline \hline \rule{0pt}{2.5ex}\ \ \ \text {House}(\%) \ \ \ \rule[-1ex]{0pt}{0pt}& \text {Apartment}(\%) \\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 30 \\
\hline \hline \rule{0pt}{2.5ex}40 \rule[-1ex]{0pt}{0pt}& 50 \\
\hline \hline \rule{0pt}{2.5ex}60 \rule[-1ex]{0pt}{0pt}& 20 \\
\hline \hline \rule{0pt}{2.5ex}100 \rule[-1ex]{0pt}{0pt}& 100 \\
\hline
\end{array}

Networks, GEN1 2025 VCAA 34 MC

Consider the following graph.
 

The number of bridges in this graph is

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

\(C\)

Show Worked Solution

\(\text{Bridge – an edge, if removed, will disconnect the network.}\)

\(\text{The top 3 edges satisfy this test.}\)

\(\Rightarrow C\)

Matrices, GEN1 2025 VCAA 29 MC

The female population of an animal species has been divided into four age groups (1, 2, 3 and 4), with age group 1 being the youngest.

The age groups, birth rates and survival rates are presented in the life-cycle diagram below.
 

 
Which one of the following is a Leslie matrix that corresponds with this life-cycle diagram?
 

  1. \(\begin{bmatrix}0 & 2.1 & 4.6 & 1.8 \\ 0.9 & 0 & 0 & 0 \\ 0 & 0.7 & 0 & 0 \\ 0 & 0 & 0.2 & 0\end{bmatrix}\)
     
  2. \(\begin{bmatrix}0 & 0.9 & 0.7 & 0.2 \\ 2.1 & 0 & 0 & 0 \\ 0 & 4.6 & 0 & 0 \\ 0 & 0 & 1.8 & 0\end{bmatrix}\)
     
  3. \(\begin{bmatrix}2.1 & 4.6 & 1.8 & 0 \\ 0.9 & 0 & 0 & 0 \\ 0 & 0.7 & 0 & 0 \\ 0 & 0 & 0.2 & 0\end{bmatrix}\)
     
  4. \(\begin{bmatrix}2.1 & 4.6 & 1.8 & 0.8 \\ 0 & 0.9 & 0 & 0 \\ 0 & 0 & 0.7 & 0 \\ 0 & 0 & 0 & 0.2\end{bmatrix}\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Age group 1 has no birthrate \(\Rightarrow  \ e_{11}=0\) (eliminate C, D)}\)

\(\text{Only option A has birth/death rates in the correct place.}\)

\(\Rightarrow A\)

Matrices, GEN1 2025 VCAA 27 MC

Consider the matrix \(E\) where

\begin{align*}
E=\begin{bmatrix}
m & -9 \\
4 & n
\end{bmatrix}
\end{align*}

For the inverse of \(E\) to exist, the values of \(m\) and \(n\), respectively, cannot be

  1. \(3\) and \(12\)
  2. \(12\) and \(3\)
  3. \(3\) and \(-12\)
  4. \(-3\) and \(-12\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Inverse exists}\ \ \Rightarrow\ \ \operatorname{det}\,E \neq 0\)

\(\operatorname{det}\,E=m \times n-(-9 \times 4)=mn+36\)

\(\text {Only option C gives} \ \ mn=-36\)

\(\Rightarrow C\)

Matrices, GEN1 2025 VCAA 26 MC

Consider the matrices \(A, B\) and \(C\) where

\begin{align*}
A=\begin{bmatrix}
2 & 0 \\
1 & 6
\end{bmatrix}, B=\begin{bmatrix}
3 \\
5
\end{bmatrix} \ \text{and}\ \  C=AB.
\end{align*}

The calculation that correctly determines element \(c_{21}\) is

  1. \(2 \times 3+0 \times 5\)
  2. \(2 \times 5+3 \times 0\)
  3. \(1 \times 3+6 \times 5\)
  4. \(1 \times 5+6 \times 3\)
Show Answers Only

\(C\)

Show Worked Solution

\(AB=\begin{bmatrix}2 & 0 \\ 1 & 6\end{bmatrix}\begin{bmatrix}3 \\ 5\end{bmatrix}=\begin{bmatrix} 6 \\ 33\end{bmatrix}\)

\(c_{12}=1 \times 3+6 \times 5=33\)

\(\Rightarrow C\)

Matrices, GEN1 2025 VCAA 25 MC

Consider the matrix \(G\) where

\begin{align*}
G=\begin{bmatrix}
0 & 1 & 0 \\
1 & 0 & 1 \\
0 & 0 & 0
\end{bmatrix}
\end{align*}

Which one of the following correctly describes matrix \(G\)?

  1. a binary matrix
  2. a permutation matrix
  3. an identity matrix
  4. a diagonal matrix
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Binary matrix – made up of 0’s and 1’s only.}\)

\(\text{Permutation matrix – single 1 in each row and column (not B).}\)

\(\text{Identity matrix – main diagonal of 1’s (not C).}\)

\(\text{Diagonal matrix – all entries not on main diagonal =0 (not D).}\)

\(\Rightarrow A\)

Data Analysis, GEN1 2025 VCAA 15 MC

The number of drinks sold daily by a juice bar, drinks, over a 10-day period is shown in the table below.
 

The four-mean smoothed number of drinks, with centring, sold on day 8 is closest to

  1. 134.25
  2. 140.0
  3. 142.75
  4. 145.75
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Calculation 1:} \ \ \dfrac{78+187+106+166}{4}=134.25\)

\(\text{Calculation 2:} \ \ \dfrac{187+106+166+124}{4}=145.75\)

\(\text{Centring:} \ \ \dfrac{134.25+145.75}{2}=140\)

\(\Rightarrow B\)

Data Analysis, GEN1 2025 VCAA 13-14 MC

The following time series graph shows the margin, in points, for a team winning all of its games in the first 18 weeks of a season. The team plays one game per week.
 

Part 1

The time series graph is smoothed using five-median smoothing.

The smoothed value for the margin in week 8, in points, is

  1. 32
  2. 35
  3. 38
  4. 41

 
Part 2

Which one of the following options best describes the qualitative features of the time series graph above?

  1. decreasing trend only
  2. irregular fluctuations only
  3. structural change only
  4. decreasing trend with irregular fluctuations
Show Answers Only

\(\text{Part 1:}\ C\)

\(\text{Part 2:}\ D\)

Show Worked Solution

\(\text{Part 1}\)

\(\text{Five values with week 8 in the middle:} \ 50,38,32,35,41\)

\(\text{Ranking values (above):} \ 32,35,38,41,50\)

\(\text{Median}=38\)

\(\Rightarrow C\)
 

\(\text{Part 2}\)

\(\text{Qualitative features of graph:}\)

\(\text{Decreasing trend (eliminate B and C)}\)

\(\text{Irregular fluctuations}\)

\(\Rightarrow D\)

Data Analysis, GEN1 2025 VCAA 11-12 MC

The table below shows the life expectancy in years, life, and the number of doctors per 1000 people, doctors, for a sample of 10 countries in 2024. A scatterplot displaying the data is also shown.
 

 

Part 1

A logarithmic (base 10) transformation was applied to the variable life.

With \(\log _{10}(\textit{life})\) as the response variable, the equation of the least squares line fitted to the transformed data is closest to

  1. \(\log _{10}(life)=1.57+0.0123 \times doctors\) 
  2. \(\log _{10}(life)=1.63+0.0326 \times doctors\)
  3. \(\log _{10}(life)=1.79+0.0383 \times doctors\)
  4. \(\log _{10}(life)=1.85+0.0403 \times doctors\)

 
Part 2

A squared transformation was applied to the variable \(doctors\).

The equation of the least squares line fitted to this transformed data is of the form  \(\textit{life}=a+b \times(\textit{doctors})^2\).

Using this equation, the predicted \(life\), in years, for a country with two \(doctors\) per 1000 people is closest to

  1. 73.6
  2. 74.0
  3. 74.5
  4. 74.9
Show Answers Only

\(\text{Part 1:}\ C\)

\(\text{Part 2:}\ C\)

Show Worked Solution

\(\text{Transform data in table (use for parts 1 and 2):}\)

\begin{array} {|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \textit{doctors} \ \ \rule[-1ex]{0pt}{0pt} & \log_{10}(\textit{life}) & \textit{doctors}^2 & \quad \textit{life} \quad \\
\hline
\rule{0pt}{2.5ex} 0.21 \rule[-1ex]{0pt}{0pt} & 1.810 &0.044 &64.6\\
\hline
\rule{0pt}{2.5ex} 0.42 \rule[-1ex]{0pt}{0pt} & 1.809 &0.176 &64.4\\
\hline
\rule{0pt}{2.5ex} 0.59 \rule[-1ex]{0pt}{0pt} & 1.811 &0.348 &64.7\\
\hline
\rule{0pt}{2.5ex} 0.79 \rule[-1ex]{0pt}{0pt} & 1.814 &0.624 &65.1\\
\hline
\rule{0pt}{2.5ex} 1.12 \rule[-1ex]{0pt}{0pt} & 1.819 &1.254 &65.9\\
\hline
\rule{0pt}{2.5ex} 1.42 \rule[-1ex]{0pt}{0pt} & 1.824 &2.016 &66.7\\
\hline
\rule{0pt}{2.5ex} 1.72 \rule[-1ex]{0pt}{0pt} & 1.836 &2.958 &68.6\\
\hline
\rule{0pt}{2.5ex} 1.77 \rule[-1ex]{0pt}{0pt} & 1.851 &3.133 &70.9\\
\hline
\rule{0pt}{2.5ex} 1.94 \rule[-1ex]{0pt}{0pt} & 1.868 &3.764 &73.8\\
\hline
\rule{0pt}{2.5ex}2.05 \rule[-1ex]{0pt}{0pt} & 1.898 &4.203 &79.1\\
\hline
\end{array}

 
\(\text{Part 1}\)

\(\text{Calculate LSRL (by CAS):}\)

\(\log _{10}(\textit{life})=1.7879 \ldots+0.03829 \ldots \times \textit{doctors}\)

\(\Rightarrow C\)
 

\(\text{Part 2}\)

\(\text{Calculate LSRL (by CAS):}\)

\(\textit{life}=63.1165+2.8419 \times \textit{doctors}^2\)
 

\(\text{Find}\ \textit{life}\ \text{when} \ \textit{doctors}=2:\)

\(\textit{life}=63.1165+2.8419 \times 2^2=74.48\ldots \ \text{years}\)

\(\Rightarrow C\)

Data Analysis, GEN1 2025 VCAA 7-8 MC

The data in the table below shows the preferred car colour for a sample of female and male car buyers.
 

Part 1

From this table, the percentage of female car buyers whose preferred car colour is silver is closest to

  1. 33%
  2. 40%
  3. 42%
  4. 57%

 
Part 2

Which one of the following is the most appropriate way to graphically display the data shown in the table above?

  1. a histogram
  2. a back-to-back stem plot
  3. parallel boxplots
  4. a segmented bar chart
Show Answers Only

\(\text{Part 1:}\ B\)

\(\text{Part 2:}\ D\)

Show Worked Solution

\(\text{Part 1}\ \)

\(\text{Total females}\ =28+42+35=105\)

\(\text{Females who prefer silver}\ = 42\)

\(\text{Percentage}\ = \dfrac{42}{105}=0.40\)

\(\Rightarrow B\)
 

\(\text{Part 2}\)

\(\text{Most appropriate graphical display is a segmented bar chart.}\)

\(\text{Each bar = a gender with colours represented by segments in each bar.}\)

\(\Rightarrow D\)

Data Analysis, GEN1 2025 VCAA 6 MC

The following information relating to life expectancy comes from a sample of nations in the Oceania region.

  • The first quartile is 74.9 years.
  • The third quartile is 78.5 years.
  • The lowest five values recorded are 68.5, 68.6, 69.0, 70.1 and 74.8 years.

How many outliers would be displayed at the lower end of a boxplot showing this sample of Oceania data?

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

\(C\)

Show Worked Solution

\(IQR=78.5-74.9=3.6\)

\(\text{Lower fence}\ = 74.9-1.5 \times 3.6=69.5\)

\(\text{68.5, 68.6 and 69.0 are all less than the lower fence.}\)

\(\Rightarrow C\)

Data Analysis, GEN1 2025 VCAA 5 MC

The heights of females aged between 16 and 18 years within a population are normally distributed.

Analysis of the heights of this group of females showed that:

  • 2.5% of the heights were greater than 178.9 cm
  • 16% of the heights were less than 157.6 cm.

Using the 68-95-99.7% rule, the mean and standard deviation of the heights of these females are respectively

  1. 150.5 and 21.3
  2. 154.9 and 7.1
  3. 164.7 and 7.1
  4. 171.8 and 14.2
Show Answers Only

\(C\)

Show Worked Solution
\(\bar{x}+2z\) \(=178.9\ \ …\ (1)\)  
\(\bar{x}-z\) \(=157.6\ …\ (2)\)  

 
\(\text{Subtract}\ (1)-(2):\)

\(3z=21.3\ \ \Rightarrow\ \ z=7.1\)

\(\text{Substitute}\ \ z=7.1\ \ \text{into (2):}\)

\(\bar{x}-7.1=157.6\ \ \Rightarrow\ \ \bar{x}=164.7\)

\(\Rightarrow C\)

Data Analysis, GEN1 2025 VCAA 3 MC

The boxplots below show the life expectancy, in years, for a sample of countries from two different continents (Sample H and Sample T).
 

Which one of the following statements is correct?

  1. The interquartile range for Sample T is greater than the interquartile range for Sample H.
  2. The median for Sample T is more than 10 years greater than the median for Sample H.
  3. The third quartile for Sample H is greater than the first quartile for Sample T.
  4. Life expectancy in all Sample T countries exceeds the median life expectancy in Sample H.
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Options A to C are all demonstrably incorrect.}\)

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

\(\text{Sample T’s low is higher than Sample H’s median.}\ \checkmark \)

\(\Rightarrow D\)

Data Analysis, GEN1 2025 VCAA 1-2 MC

At a fruit shop, customers can buy avocados in bags.

The bag size ranges from one to six avocados per bag.

The histogram below shows the number of customers who bought each bag size on a particular day.
 

Part 1

The median bag size bought by customers on the day was

  1. 2
  2. 3
  3. 3.5
  4. 4


Part 2

The total number of avocados sold in bags was

  1. 39
  2. 134
  3. 136
  4. 138
Show Answers Only

\(\text{Part 1:}\ D\)

\(\text{Part 2:}\ D\)

Show Worked Solution

\(\text{Part 1}\)

\(\text{Sum of columns}\ = 4+11+3+9+5+7=39\)

\(\text{Median = 20th value which is in the 4th column (bag size 4).}\)

\(\Rightarrow D\)
 

\(\text{Part 2}\)

\(\text{Total number of avocados}\)

\(=1 \times 4 + 2 \times 11+3 \times 3+4 \times 9+5 \times 5+6 \times 7 =138\)

\(\Rightarrow D\)

Calculus, 2ADV C1 EQ-Bank 12

A block of ice is melting. The mass \(M\) kilograms of the ice block remaining at time \(t\) hours after it begins to melt is given by  \(M(t)=50(12-3t)^2, 0 \leqslant t \leqslant 4\).

  1. Find the rate of change of the ice block's mass at any time \(t\).   (1 mark)

  2. How long does it take for the ice block to completely melt?   (1 mark)

  3. At what time is the ice melting at a rate of 2100 kilograms per hour?   (2 marks)

Show Answers Only

a.    \(\dfrac{dM}{dt}=-300(12-3t)\)

b.    \(4\ \text{hours}\)

c.    \(t=\dfrac{5}{3}\ \text{hours}\)

Show Worked Solution

a.    \(M(t)=50(12-3t)^2\)

\(\dfrac{dM}{dt}=50 \times 2 \times (-3) \times(12-3t)=-300(12-3t)\)
 

b.    \(\text{Find}\ t\ \text{when}\ \ M(t)=0:\)

\(50(12-3t)^2=0 \ \Rightarrow \ t=4\)

\(\text{Ice block is completely melted at} \ \ t=4 \ \ \text {hours}\)
 

c.    \(\text{Find}\ t \ \text{when}\ \ \dfrac{d M}{d t}=-2100:\)

\(-300(12-3t)\) \(=-2100\)
\(12-3t\) \(=7\)
\(-3t\) \(=-5\)
\(t\) \(=\dfrac{5}{3}\ \text{hours}\)

Calculus, 2ADV C1 EQ-Bank 11

An oil slick on the surface of water forms a circular shape. The radius \(r\) metres of the oil slick is increasing according to the formula  \(r(t)=3 \sqrt{t}\), where \(t\) is the time in minutes after the oil begins to spread,  \(t \geqslant 0\).

  1. Find the rate at which the radius is increasing at any time \(t\).   (1 mark)

  2. At what rate is the area of the oil slick increasing when \(t=16\) minutes?   (1 mark)

Show Answers Only

a.    \(\dfrac{d r}{d t}=\dfrac{3}{2 \sqrt{t}}\)

b.    \(\dfrac{d r}{d t}=\dfrac{3}{8} \ \text{metres/min}\)

Show Worked Solution

a.    \(r(t)=3 \sqrt{t}\)

\(\dfrac{d r}{d t}=\dfrac{1}{2} \times 3 \times t^{-\tfrac{1}{2}}=\dfrac{3}{2 \sqrt{t}}\)
 

b.    \(\text{Find} \ \dfrac{d r}{d t} \ \text{when} \ \ t=16:\)

\(\dfrac{d r}{d t}=\dfrac{3}{2 \times \sqrt{16}}=\dfrac{3}{8} \ \text{metres/min}\)

Calculus, 2ADV C1 EQ-Bank 13

A cylindrical water tank is being filled. The volume \(V\) litres of water in the tank at time \(t\) minutes after filling begins is given by  \(V(t)=500 \sqrt{(2 t+1)}, t \geqslant 0\).

  1. At what rate is water entering the tank at any time \(t\) ?   (1 mark)

  2. Find the rate at which the tank is being filled when \(t=12\) minutes.   (1 mark)

  3. At what time is the water flowing into the tank at a rate of 125 litres per minute?   (2 marks)

Show Answers Only

a.    \(\dfrac{dV}{dt}= \dfrac{500}{\sqrt{2t+1}} \)

b.    \(100\ \text{L/min} \)

c.    \(\dfrac{15}{2}\ \text{mins}\)

Show Worked Solution

a.    \(V(t)=500(2t+1)^{\frac{1}{2}}\)

\(\dfrac{dV}{dt}=\dfrac{1}{2} \times 2 \times 500(2t+1)^{-\frac{1}{2}} = \dfrac{500}{\sqrt{2t+1}} \)
 

b.    \(\text{When}\ \ t=12:\)

\(\dfrac{dV}{dt}= \dfrac{500}{\sqrt{25}} = 100\ \text{L/min} \)
 

c.    \(\text{Find}\ t\ \text{when}\ \dfrac{dV}{dt}=125:\)

\(125\) \(=\dfrac{500}{\sqrt{2t+1}}\)  
\(\sqrt{2t+1}\) \(=4\)  
\(2t+1\) \(=16\)  
\(t\) \(=\dfrac{15}{2}\ \text{mins}\)  

Financial Maths, STD2 EQ-Bank 29

Priya works as a sales representative and earns a base wage plus commission. A spreadsheet is used to calculate her weekly earnings.

Total weekly earnings = Base wage + Total sales \(\times\) Commission rate

A spreadsheet showing Priya's weekly earnings is shown.
  

 

  1. Write down the formula used in cell B9, using appropriate grid references.   (1 mark)

  2. In the following week, Priya earned total weekly earnings of $1437.50. Her base wage and commission rate remained unchanged. Calculate Priya's total sales for that week.   (2 marks)

  3. Priya's employer increases her commission rate to 4.2% but keeps her base wage the same. If Priya makes $22 000 in sales, calculate her new total weekly earnings.   (2 marks)

Show Answers Only

a.   \(=\text{B4}+\text{B5}^*\text{B6}/100\)

b.    \($22\, 500\)

c.    \($1574\)

Show Worked Solution

a.   \(\text{Total weekly earnings} = \text{Base wage}+\text{Total sales} \times \text{Commission rate}\)

\(\therefore\ \text{Formula:}\ =\text{B4}+\text{B5}^*\text{B6}/100\)
 

b.   \(\text{Total weekly earnings} = \text{Base wage}+\text{Total sales} \times \text{Commission rate}\)

\(\text{Let the Total sales}=S\)

\(1437.50\) \(=650+S\times \dfrac{3.5}{100}\)
\(787.50\) \(=S\times \dfrac{3.5}{100}\)
\(S\) \(=\dfrac{787.50\times 100}{3.5}=$22\,500\)

 
\(\text{The amount of Priya’s total sales was \$22 500.}\)
 

c.    \(\text{Base wage}=650\)

\(\text{Total sales}=$22\,000\)

\(\text{New commission rate}=4.2\%\)

\(\text{Total weekly earnings}\) \(=650+22\,000\times \dfrac{4.2}{100}\)
  \(=650+924=$1574\)

Financial Maths, STD2 EQ-Bank 21

Liam works in a factory assembling electronic components and is paid on a piecework basis. A spreadsheet is used to calculate his weekly earnings.

Weekly earnings = Number of units completed \(\times\) Rate per unit

A spreadsheet showing Liam's earnings for one week is shown.
  

  1. Write down the formula used in cell B8, using appropriate grid references.   (1 mark)

  2. In the following week, Liam earned $2036.25. The rate per unit remained at $3.75. Calculate the number of units Liam completed that week.   (2 marks)

Show Answers Only

a.   \(=\text{B4}^*\text{B5}\)

b.    \(\text{Number of units completed}=543\)

Show Worked Solution

a.   \(\text{Weekly earnings} = \text{Number of units completed} \times \text{Rate per unit}\)

\(\text{Formula:}\ =\text{B4}^*\text{B5}\)
 

b.   \(\text{Weekly earnings} = \text{Number of units completed} \times \text{Rate per unit}\)

\(\text{Let the Weekly earnings}=E\)

\(2036.25\) \(=E\times 3.75 \)
\(E\) \(=\dfrac{2036.25}{3.75}=543\)

Financial Maths, STD2 EQ-Bank 26

Maria works as a freelance writer and earns income through royalties. A spreadsheet is used to calculate her monthly royalty earnings.

Royalties = Number of books sold \( \times \) Royalty rate per book

A spreadsheet showing Maria's royalty earnings is shown.
  

  1. Write down the formula used in cell B8, using appropriate grid references.   (1 mark)

  2. In April, Maria's total royalty earnings were $8960. Her royalty rate remained at $2.85 per book. How many books did Maria sell in April?   (2 marks)

Show Answers Only

a.   \(=\text{B4}^*\text{B5}\)

b.    \(\text{Number of books}=3144\)

Show Worked Solution

a.   \(\text{Total royalty earnings} = \text{Number of books sold} \times \text{Royalty rate per book}\)

\(\text{Formula:}\ =\text{B4}^*\text{B5}\)
 

b.   \(\text{Total royalty earnings} = \text{Number of books sold} \times \text{Royalty rate per book}\)

\(\text{Let the Number of books sold}=N\)

\(8960.40\) \(=N\times 2.85 \)
\(N\) \(=\dfrac{8960.40}{2.85}=3144\)

Measurement, STD2 EQ-Bank 23

2UG-2005-25b

Use Pythagoras’ theorem to show that `ΔABC` is a right-angled triangle.   (1 mark)

Show Answers Only

`ΔABC\ text(is right-angled if)\ \ a^2 + b^2 = c^2:`

`a^2 + b^2= 5^2 + 12^2= 169= 13^2= c^2`

Show Worked Solution

`ΔABC\ text(is right-angled if)\ \ a^2 + b^2 = c^2:`

`a^2 + b^2= 5^2 + 12^2= 169= 13^2= c^2`

Financial Maths, STD2 EQ-Bank 31

Chen earns an annual salary of $72 800. He is entitled to four weeks annual leave with 17.5% leave loading. A spreadsheet is used to calculate his total holiday pay.

Total holiday pay = 4 × weekly wage + 4 × weekly wage × 17.5%

A spreadsheet showing Chen's holiday pay calculation is shown.

  1. What value from the question should be entered in cell B5?   (1 mark)

  2. Write down the formula used in cell B9 to calculate the weekly wage, using appropriate grid references.   (2 marks)
  3. Verify the amount of Chen's total holiday pay using calculations.   (2 marks)
Show Answers Only

a.   \(4\)

b.   \(=\text{B4}/52\)

c.    \(\text{Total holiday pay}=\text{weekly wage}\times 4 +\ \text{weekly wage}\times 4\ \times 17.5\%\)

\(\text{Total holiday pay}=1400\times 4+1400\times 4\times\dfrac{17.5}{100}=5600+980=$6580\)

Show Worked Solution

a.    \(\text{Chen gets 4 weeks leave }\rightarrow\ 4\)
 

b.    \(\text{Weekly wage }=\dfrac{\text{Annual salary}}{52}\)

\(\text{Formula: }=\text{B4}/52\)
 

c.    \(\text{Total holiday pay}=\text{weekly wage}\times 4 +\ \text{weekly wage}\times 4\ \times 17.5\%\)

\(\text{Total holiday pay}=1400\times 4+1400\times 4\times\dfrac{17.5}{100}=5600+980=$6580\)

Algebra, STD2 EQ-Bank 29

The formula below is used to estimate the number of hours you must wait before your blood alcohol content (BAC) will return to zero after consuming alcohol.

\(\text{Number of hours}\ =\dfrac{\text{BAC}}{0.015}\)

The spreadsheet below has been created by Ben so his 21st birthday attendees can monitor their alcohol consumption if they intend to drive, given their BAC reading. 

  1. By using appropriate grid references, write down a formula that could appear in cell B5.   (2 marks)

  2. Ben's friend Ryan's reading reflects that he will have to wait 11 hours for his BAC to return to zero. Using the formula, calculate the value Ryan entered into cell B3.   (2 marks)

Show Answers Only

a.    \(=\text{B3}/0.015\)

b.    \(0.165\)

Show Worked Solution

a.    \(=\text{B3}/0.015\)

b.    \(11\) \(=\dfrac{\text{BAC}}{0.015}\)
  \(\text{BAC}\) \(=11\times 0.015=0.165\)

  
\(\text{Ryan entered 0.165 into cell B3.}\)

Financial Maths, STD1 EQ-Bank 24

A company uses the spreadsheet below to calculate the fortnightly pay, after tax, of its employees.
 

  1. Write down the formula that was used in cell D5, using appropriate grid references.   (1 mark)
  2. Hence, calculate Kim's fortnightly pay.   (2 marks)
Show Answers Only

a.  \(=\text{B6}-\text{C6}\)

b.    \(\$3140.85\)

Show Worked Solution

a.    \(\text{Formula for Kim’s Salary after tax:}\)

\(=\text{B5}-\text{C5}\)
 

b.    \(\text{Kim’s salary after tax}\ = \$81\,662\)

\(\text{Kim’s fortnightly pay}\ = \dfrac{81\,662}{26}=\$3140.85\)

Financial Maths, STD2 EQ-Bank 17

The table shows the income tax rate for Australian residents for the 2024-2025 financial year.

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\textit{Taxable income} \rule[-1ex]{0pt}{0pt}& \textit{Tax on this income} \\
\hline
\rule{0pt}{2.5ex}0-\$18\,200 \rule[-1ex]{0pt}{0pt}& \text{Nil} \\
\hline
\rule{0pt}{2.5ex}\$18 \, 201-\$45\,000 \rule[-1ex]{0pt}{0pt}& \text{16 cents for each \$1 over \$18 200} \\
\hline
\rule{0pt}{2.5ex}\$45\,001-\$135\,000 \rule[-1ex]{0pt}{0pt}& \$4288 \text{ plus 30 cents for each \$1 over \$45 000} \\
\hline
\rule{0pt}{2.5ex}\$135\,001-\$190\,000 \rule[-1ex]{0pt}{0pt}& \$31 \, 288 \text{ plus 37 cents for each \$1 over \$135 000} \\
\hline
\rule{0pt}{2.5ex}\$190\,001 \text{ and over} \rule[-1ex]{0pt}{0pt}& \$51 \, 638 \text{ plus 45 cents for each \$1 over \$190 000} \\
\hline
\end{array}

A company's spreadsheet was created that calculates its employees' after tax fortnightly pay, based on the table and excluding the Medicare levy.

  1. Write down the formula that was used in cell D6, using appropriate grid references.   (1 mark)
  2. Write down the formula that was used in cell C5, using appropriate grid references.   (1 mark)
  3. Hence, calculate Greg's fortnightly pay.   (2 marks)
Show Answers Only

a.  \(=\text{B6}-\text{C6}\)

b.  \(=4288+(\text{B5}-45000)^*0.30 \)

c.    \(\$3140.85\)

Show Worked Solution

a.    \(\text{Formula for Ian’s Salary after tax:}\)

\(=\text{B6}-\text{C6}\)
 

b.    \(\text{Formula for Greg’s estimated tax:}\)

\(=4288+(\text{B5}-45000)^*0.30 \)
 

c.    \(\text{Greg’s estimated tax}\ =4288+(103\,500-45\,000) \times 0.30=\$21\,838\)

\(\text{Greg’s salary after tax}\ = 103\,500-21\,838=\$81\,662\)

\(\text{Greg’s fornightly pay}\ = \dfrac{81\,662}{26}=\$3140.85\)

Algebra, STD2 EQ-Bank 23

Sharon drinks three glasses of chardonnay over a 180-minute period, each glass containing 1.6 standard drinks.

Sharon weighs 78 kilograms, and her blood alcohol content (BAC) at the end of this period can be calculated using the following formula:

\(\text{BAC}_{\text {female }}=\dfrac{10 N-7.5 H }{5.5 M}\)

where \(N\) = number of standard drinks consumed
\(H\) = the number of hours drinking
\(M\) = the person's mass in kilograms

 
The spreadsheet below can be used to calculate Sharon's \(\text{BAC}\).
 

  1. What value should be input into cell B5.   (1 mark)

  2. Write down the formula that has been used in cell E4, using appropriate grid references.   (2 marks)

Show Answers Only

a.    \(\text{Cell B5 value}=3\)

b.  \(=\left(10^* \text{B4}-7.5^* \text{B5}\right) /(5.5^* \text{B6})\)

Show Worked Solution

a.    \(\text{180 minutes}\ =\ \text{3 hours}\)

\(\therefore \ \text{Cell B5 value}=3\)
 

b.  \(=\left(10^* \text{B4}-7.5^* \text{B5}\right) /(5.5^* \text{B6})\)

Algebra, STD2 EQ-Bank 25

A fitness app calculates daily calorie requirements using the formula below.

Daily calories = Basal metabolic rate + Calorie burn rate per hour \( \times \) Hours of activity

The spreadsheet below has been used to calculate Jamal's daily calorie requirements when he has had 6 hours of activity.
  

  1. Write down the formula used in cell B9, using appropriate grid references.   (1 mark)

  2. Jamal increases his hours of activity to 8 hours per day, while his basal metabolic rate and calorie burn rate remain the same.

    What will be Jamal's new daily calorie requirement?   (1 mark)

Show Answers Only

a.    \(=\text{B4}+ \text{B5} \ ^* \ \text{B6}\)

b.    \(2770\ \text{calories}\)

Show Worked Solution

a.   \(=\text{B4}+ \text{B5} \ ^* \ \text{B6}\)

b.    \(\text{Daily calories}=1650+140\times 8=2770\)

\(\therefore\ \text{Jamal’s new daily calorie requirement is}\ 2770.\)

Algebra, STD2 EQ-Bank 24

QuickPrint Copy Centre charges for printing services using the formula below.

Total cost = Setup fee + Cost per page \( \times \) Number of pages

A spreadsheet used to calculate the total cost is shown.

  1. Write down the formula used in cell E3, using appropriate grid references.   (1 mark)

  2. QuickPrint increases their cost per page to \$0.42, but keeps the setup fee unchanged. Aisha needs to print 120 pages.

    How much more will Aisha pay compared to the original pricing shown in the spreadsheet?   (2 marks)

Show Answers Only

a.    \(=\text{B3}+ \text{B4} \ ^* \ \text{B5}\)

b.    \($14.40\)

Show Worked Solution

a.   \(=\text{B3}+ \text{B4} \ ^* \ \text{B5}\)
 

b.   \(\text{Original cost from spreadsheet}:\ $44.50\)

\(\text{At increased rate}:\)

\(\text{Total cost}\ =8.50+0.42\times 120=$58.90\)

\(\text{Additional amount}\ =58.90-44.50=$14.40\)

Algebra, STD2 EQ-Bank 23

Green Thumb Landscaping charges for their lawn mowing service based on the size of the lawn.

They use the formula below to calculate the cost of each service.

Total cost = Call-out fee + Cost per square metre \( \times \) Area of lawn

The spreadsheet they provide to their clients is included below.

  1. Write down the formula that has been used in cell E4, using appropriate grid references.   (1 mark)

  2. Miguel has a lawn with a different area. The call-out fee and cost per square metre remain the same. When Miguel's lawn area is entered into the spreadsheet, the total cost shown in cell E4 becomes \$153.00.

    What is the area of Miguel's lawn?   (2 marks)

Show Answers Only

a.    \(=\text{B4}+ \text{B5} \ ^* \ \text{B6}\)

b.    \(\text{60 m}^2\)

Show Worked Solution

a.    \(=\text{B4}+ \text{B5} \ ^* \ \text{B6}\)
 

b.     \(153\) \(=45+1.8A\)
  \(108\) \(=1.8A\)
  \(A\) \(=\dfrac{108}{1.8}=60\)

 
\(\therefore\ \text{The area of Miguel’s lawn is 60 m}^2.\)

Algebra, STD2 EQ-Bank 24

It is known that a quantity \(N\) varies directly with another quantity \(Q\).

The relationship can be modelled by the equation  \(N= k \times Q\), where \(k\) is a constant.

If \(N = 18\)  when  \(Q=4:\)

  1. Show the value of  \(k=4.5\).   (1 mark)

  2. Hence find the value of \(Q\) when  \(N=63\).  (1 mark)

Show Answers Only

a.    \(\text{Using}\ \ N=k \times Q:\)

\(18= k \times 4\ \ \Rightarrow\ \ k=\dfrac{18}{4}=4.5\)

b.    \(Q=14\)

Show Worked Solution

a.    \(\text{Using}\ \ N=k \times Q:\)

\(18= k \times 4\ \ \Rightarrow\ \ k=\dfrac{18}{4}=4.5\)
 

b.    \(\text{Find}\ Q\ \text{when}\ \ N=63:\)

\(63\) \(=4.5 \times Q\)  
\(Q\) \(=\dfrac{63}{4.5}=14\)  

Polynomials, EXT1 EQ-Bank 16

The polynomial  \(R(x)=x^3+p x^2+q x+6\)  has a double zero at  \(x=-1\)  and a zero at  \(x=s\).

Find the values of \(p, q\) and \(s\).   (3 marks)

Show Answers Only

\(s=-6, \ p=8, \ q=13\)

Show Worked Solution

\(R(x)=x^3+p x^2+q x+6\)

\(R(x)\ \text{is monic with a zero at} \ s \ \text{and double zero at}\ -1:\)

\(R(x)\) \(=(x+1)^2(x-s)\)
  \(=\left(x^2+2 x+1\right)(x-s)\)
  \(=x^3+2 x^2+x-s x^2-2 s x-s\)
  \(=x^3+(2-s) x^2+(1-2 s) x-s\)

 

\(\text{Equating coefficients:}\)

\(-s=6 \ \Rightarrow \ s=-6\)

\(p=2-(-6)=8\)

\(q=1-2(-6)=13\)

Polynomials, EXT1 EQ-Bank 19

A polynomial has the equation

\(Q(x)=(x+1)^2(x-2)\left(x^2+2 x-8\right)\)

  1. Express  \(Q(x)\)  as a product of linear factors and determine the multiplicity of each of its roots.   (2 marks)

  2. Hence, without using calculus, draw a sketch of \(y=Q(x)\), showing all  \(x\)-intercepts.   (2 marks)

Show Answers Only

a.    \(Q(x) = (x+1)^2(x-2)\left(x^2+2 x-8\right) \)

\(\text{Roots:}\)

\(x=-1 \ \ \text{(multiplicity 2)}\)

\(x=2 \ \ \text{(multiplicity 2)}\)

\(x=4 \ \ \text{(multiplicity 1)}\)
 

b.

Show Worked Solution
a.     \(Q(x)\) \(=(x+1)^2(x-2)\left(x^2+2 x-8\right)\)
    \(=(x+1)^2(x-2)(x-2)(x+4)\)
    \(=(x+1)^2(x-2)^2(x-4)\)

 

\(\text{Roots:}\)

\(x=-1 \ \ \text{(multiplicity 2)}\)

\(x=2 \ \ \text{(multiplicity 2)}\)

\(x=4 \ \ \text{(multiplicity 1)}\)
 

b.    \(Q(x) \ \text{degree}=5, \ \text{Leading coefficient}=1\)

\(\text{As} \ \ x \rightarrow-\infty, y \rightarrow-\infty\)

\(\text{As} \ \ x \rightarrow \infty, y \rightarrow \infty\)

\(\text{At}\ \ x=0, \ y=1^2 \times (-2)^2 \times -4 = -16\)

Polynomials, EXT1 EQ-Bank 11

A polynomial \(f(x)\) is defined by

\(f(x)=-3x^3+27x^2+12x-1\)

Explain what happens to \(f(x)\) as  \(x \rightarrow-\infty\).   (2 marks)

Show Answers Only

\(\text{Since}\ f(x) \ \text{has degree 3:}\)

\(\text{As} \ \ x \rightarrow-\infty, x^3 \rightarrow-\infty\)

\(f(x) \ \text{has leading coefficient}=-3\)

\(\therefore\ \text{As} \ \ x \rightarrow-\infty,-3 x^3 \rightarrow \infty, \ f(x) \rightarrow \infty\)

Show Worked Solution

\(\text{Since}\ f(x) \ \text{has degree 3:}\)

\(\text{As} \ \ x \rightarrow-\infty, x^3 \rightarrow-\infty\)

\(f(x) \ \text{has leading coefficient}=-3\)

\(\therefore\ \text{As} \ \ x \rightarrow-\infty,-3 x^3 \rightarrow \infty, \ f(x) \rightarrow \infty\)

Polynomials, EXT1 EQ-Bank 15

Consider a polynomial  \(y=(x+3)^2(2-x) \cdot Q(x)\)

where  \(Q(x)=6-x-x^2\)

Explain what happens to \(y\) as  \(x \rightarrow-\infty\).   (2 marks)

Show Answers Only
\(y\) \(=(x+3)^2(2-x)\left(6-x-x^2\right)\)
  \(=(x+3)^2(2-x)(x+3)(2-x)\)
  \(=(x+3)^3(2-x)^2\)

 

\(\text{Degree\(=5\), Leading co-efficient\(=1\)}\)

\(\therefore \text{As} \ \ x \rightarrow-\infty, x^5 \rightarrow-\infty, y \rightarrow-\infty\)

Show Worked Solution
\(y\) \(=(x+3)^2(2-x)\left(6-x-x^2\right)\)
  \(=(x+3)^2(2-x)(x+3)(2-x)\)
  \(=(x+3)^3(2-x)^2\)

 

\(\text{Degree\(=5\), Leading co-efficient\(=1\)}\)

\(\therefore \text{As} \ \ x \rightarrow-\infty, x^5 \rightarrow-\infty, y \rightarrow-\infty\)

Polynomials, EXT1 EQ-Bank 18

Consider the function  \(P(x)=(x-1)^2(x+2)\left(x^2+3 x-4\right)\)

  1. By expressing \(P(x)\) as a product of its linear factors, identify its zeroes and the multiplicity of each zero.   (2 marks)

  2. Without using calculus, draw a sketch of  \(y=P(x)\)  showing any \(x\)-intercepts.   (2 marks)

Show Answers Only

a.    \(P(x)=(x-1)^3(x+2)(x+4)\)

\(\text{Roots:}\)

\(x=-2 \ \ \text{(multiplicity 1)}\)

\(x=-4 \ \ \text{(multiplicity 1)}\)

\(x=1 \ \ \text{(multiplicity 3)}\)
 

b.  

Show Worked Solution
a.     \(P(x)\) \(=(x-1)^2(x+2)\left(x^2+3 x-4\right)\)
    \(=(x-1)^2(x+2)(x-1)(x+4)\)
    \(=(x-1)^3(x+2)(x+4)\)

 
\(\text{Roots:}\)

\(x=-2 \ \ \text{(multiplicity 1)}\)

\(x=-4 \ \ \text{(multiplicity 1)}\)

\(x=1 \ \ \text{(multiplicity 3)}\)
 

b.    \(\text{Degree} \ P(x)=5, \ \ \text {Leading coefficient }=1\)

\(\text{As} \ \ x \rightarrow \infty, \ y \rightarrow \infty\)

\(\text{As} \ \ x \rightarrow -\infty, \ y \rightarrow -\infty\)

\(\text{At} \ \ x=0, y=-8\)
 

HMS, TIP 2025 HSC 31aii

How can an athlete ensure their training procedures are safe? In your answer, refer to ONE training method.  ( 5 marks)

Show Answers Only

Training Method: Plyometrics

  • Athletes must complete comprehensive warm-ups before plyometric sessions. This occurs because muscles and connective tissues require increased temperature and blood flow before explosive movements to prevent strain injuries.
  • For example, performing 10 minutes of dynamic stretching and light jogging prepares joints for impact forces. This reduces injury risk during box jumps and depth jumps.
  • Proper landing technique ensures safety during plyometric exercises. This happens when athletes maintain correct body positioning with bent knees and controlled movements, which prevents excessive joint stress.
  • Progressive overload principles must be applied carefully. This means gradually increasing jump height and repetitions over weeks, allowing tissues to adapt without overuse injuries.
  • Training on appropriate surfaces like gym mats or grass protects joints from impact forces. Consequently, athletes avoid stress fractures and tendon damage during high-intensity explosive training.
Show Worked Solution

Training Method: Plyometrics

  • Athletes must complete comprehensive warm-ups before plyometric sessions. This occurs because muscles and connective tissues require increased temperature and blood flow before explosive movements to prevent strain injuries.
  • For example, performing 10 minutes of dynamic stretching and light jogging prepares joints for impact forces. This reduces injury risk during box jumps and depth jumps.
  • Proper landing technique ensures safety during plyometric exercises. This happens when athletes maintain correct body positioning with bent knees and controlled movements, which prevents excessive joint stress.
  • Progressive overload principles must be applied carefully. This means gradually increasing jump height and repetitions over weeks, allowing tissues to adapt without overuse injuries.
  • Training on appropriate surfaces like gym mats or grass protects joints from impact forces. Consequently, athletes avoid stress fractures and tendon damage during high-intensity explosive training.

Polynomials, EXT1 EQ-Bank 3 MC

Which of the following best represents the graph of  \(y=-5 x(x-2)(3-x)\)?
 

Show Answers Only

\(C\)

Show Worked Solution

\(\text{By elimination:}\)

\(\text{Degree = 3,  Leading co-efficient}\ = 5\)

\(\text{As}\ \ x \rightarrow \infty,\ \ y \rightarrow \infty\ \text{(eliminate A and B)}\)

\(\text{When}\ x=1:\)

\(y=-5(-1)(2)=10>0\ \ \text{(eliminate D)}\)

\(\Rightarrow C\)

Polynomials, EXT1 EQ-Bank 14

Consider the polynomial \(P(x)=x(3-x)^3\).

  1. State the degree of the polynomial and identify the leading coefficient.   (1 mark)

  2. Explain what happens to \(y\) as  \(x \rightarrow \pm \infty\).   (1 mark)

  3. Without using calculus, sketch \(P(x)\) showing its general form and any \(x\)-intercepts.   (2 marks)

Show Answers Only

a.    \(\text{Degree}\ P(x)=4\)

\(\text{Leading co-efficient}=-1\)
 

b.    \(\text{As} \ \ x \rightarrow-\infty,-x^4 \rightarrow-\infty, \ y \rightarrow-\infty\).

\(\text{As} \ \ x \rightarrow \infty,-x^4 \rightarrow-\infty, \ y \rightarrow-\infty\).
 

c.    \(P(x)\ \text{has zeroes at}\ \ x=0, 3:\)


       

Show Worked Solution

a.    \(\text{Degree}\ P(x)=4\)

\(\text{Leading co-efficient}=-1\)
 

b.    \(\text{As} \ \ x \rightarrow-\infty,-x^4 \rightarrow-\infty, \ y \rightarrow-\infty\).

\(\text{As} \ \ x \rightarrow \infty,-x^4 \rightarrow-\infty, \ y \rightarrow-\infty\).
 

c.    \(P(x)\ \text{has zeroes at}\ \ x=0, 3:\)


       

Polynomials, EXT1 EQ-Bank 12

Consider a polynomial  \(y=-2 x^5-26 x^4-x+1\).

With reference to the leading term, explain what happens to \(y\) as  \(x \rightarrow-\infty\).   (2 marks)

Show Answers Only

\(\text{As}\ \ x \rightarrow-\infty,\ \ x^{5}\ \rightarrow-\infty\)

\(\text{Since the leading coefficient \((-2)\) is negative:}\)

\(\text{As}\ \ x \rightarrow-\infty,\ \ y \rightarrow \infty.\)

Show Worked Solution

\(\text{As}\ \ x \rightarrow-\infty,\ \ x^{5}\ \rightarrow-\infty\)

\(\text{Since the leading coefficient \((-2)\) is negative:}\)

\(\text{As}\ \ x \rightarrow-\infty,\ \ y \rightarrow \infty.\)

Polynomials, EXT1 EQ-Bank 20

A polynomial has the equation

\(P(x)=(x-1)(x-3)(x+2)^2\left(x^2-x-6\right)\).

  1. By expressing \(P(x)\) as a product of its linear factors, determine the multiplicity of each of the roots of  \(P(x)=0\).   (2 marks)

  2. Hence, without using calculus, draw a sketch of  \(y=P(x)\)  showing all \(x\)-intercepts.   (2 marks)

Show Answers Only
a.     \(P(x)\) \(=(x-1)(x-3)(x+2)^2\left(x^2-x-6\right)\)
    \(=(x-1)(x-3)(x+2)^2(x-3)(x+2)\)
    \(=(x-1)(x-3)^2(x+2)^3\)

 

\(\text{Roots:}\)

\(x=1\ \text{(multiplicity 1)}\)

\(x=-2\ \text{(multiplicity 3)}\)

\(x=3\ \text{(multiplicity 2)}\)
 

b.
     

Show Worked Solution
a.     \(P(x)\) \(=(x-1)(x-3)(x+2)^2\left(x^2-x-6\right)\)
    \(=(x-1)(x-3)(x+2)^2(x-3)(x+2)\)
    \(=(x-1)(x-3)^2(x+2)^3\)

 

\(\text{Roots:}\)

\(x=1\ \text{(multiplicity 1)}\)

\(x=-2\ \text{(multiplicity 3)}\)

\(x=3\ \text{(multiplicity 2)}\)
 

b.
     

Functions, EXT1 EQ-Bank 18

Consider the function  \(y=\operatorname{cosec}\,x\)  for  \(-\pi \leqslant x \leqslant \pi\).

  1. State the equations of all vertical asymptotes in the given domain.   (1 mark)

  2. Sketch the graph of  \(y=\operatorname{cosec} x\), showing all key features.   (2 marks)

     
     
Show Answers Only

a.    \(y=\operatorname{cosec}\,x=\dfrac{1}{\sin x}\)

\(\text{Asymptotes when \(\ \sin x=0 \ \) in given domain.}\)

\(\therefore \ \text{Asymptotes at} \ \ x=-\pi, 0, \pi\)
 

b.
       

Show Worked Solution

a.    \(y=\operatorname{cosec}\,x=\dfrac{1}{\sin x}\)

\(\text{Asymptotes when \(\ \sin x=0 \ \) in given domain.}\)

\(\therefore \ \text{Asymptotes at} \ \ x=-\pi, 0, \pi\)
 

b.
       

Functions, EXT1 EQ-Bank 17

  1. Sketch the graph of  \(y=\sec x\)  for  \(0 \leqslant x \leqslant 2 \pi\).
  2. In your answer, identify all asymptotes and the coordinates of any maximum and minimum turning points.   (2 marks)

     

  3. Using set notation, state the domain and range of  \(y=\sec x\).   (1 mark)

Show Answers Only

a.
   

b.    \(\text{Domain:} \ x \in\left[0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) \cup\left(\frac{3 \pi}{2}, 2 \pi\right]\)

\(\text{Range:} \ y \in(-\infty,-1] \cup[1, \infty)\)

Show Worked Solution

a.    \(\text{Draw}\ \ y=\cos\,x\ \ \text{to inform graph:}\)

 
   

\(\text{Minimum TPs:}\ (0,1), (2\pi, 1) \)

\(\text{Maximum TP:}\ (\pi, -1)\)
 

b.    \(\text{Domain:} \ x \in\left[0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) \cup\left(\frac{3 \pi}{2}, 2 \pi\right]\)

\(\text{Range:} \ y \in(-\infty,-1] \cup[1, \infty)\)

Probability, 2ADV EQ-Bank 18

A survey of 50 students found that:

  • 28 students study Mathematics (set \(M\))
  • 22 students study Physics (set \(P\) )
  • 12 students study both Mathematics and Physics.
  1. How many students study Mathematics or Physics, or both?   (1 mark)

  2. If two students are chosen at random, what is the probability that both DO NOT study either Mathematics or Physics?   (2 marks)

Show Answers Only

a.
     

\(n(M \cup P)=38 \ \text {students}\)
 

b.    \(\dfrac{66}{1225}\)

Show Worked Solution

a.
     

\(n(M \cup P)=38 \ \text {students}\)
 

b.    \(P(M \cup P)=\dfrac{38}{50}\)

\(\text{Student 1:}\ P(\overline{M \cup P})=1-\dfrac{38}{50}=\dfrac{12}{50}\)

\(\text{Student 2:} \ P(\overline{M \cup P})=\dfrac{11}{49}\)

\(\therefore P\left(\text{Both study neither }\right)=\dfrac{12}{50} \times \dfrac{11}{49}=\dfrac{66}{1225}\)

Probability, 2ADV EQ-Bank 17

A survey of 85 households asked if they subscribed to the streaming services provided by Netflix (set \(N\)), Apple TV (set \(A\)), and Stan (set \(S\)).

The survey found that 17 households had no subscription and that \(n(N)=42, n(A)=35, n(S)=28\).

The survey also found

\(n(N \cap A)=18, n(N \cap S)=15, n(A \cap S)=12\)  and  \(n(N \cap A \cap S)=8\)

  1. Complete the Venn diagram below to accurately describe the information given.   (2 marks)


     

  2. A household is chosen randomly and is found to subscribe to Apple TV. What is the probability the household also subscribes to Stan?   (2 marks)

Show Answers Only

a.
           
 

b.    \(P(S \mid A)=\dfrac{12}{35}\)

Show Worked Solution

a.
           
 

b.    \(n(A)=35, n(A \cap S)=12\)

\(P(S \mid A)=\dfrac{n(A \cap S)}{n(A)}=\dfrac{12}{35}\)

Probability, 2ADV EQ-Bank 13

Consider the universal set  \(U=\{x\) is a positive integer and  \(x \leqslant 24\}\)

Three sets are defined as

\begin{aligned}
& A=\{x \text { is a factor of } 24\} \\
& B=\{x \text { is a perfect square}\} \\
& C=\{x \text { is divisible by } 3\}
\end{aligned}

  1. List the elements of set \(A\).   (1 mark)

  2. Find \(A \cap B\).   (1 mark)

  3. Find \((A \cup B) \cap C^c\)   (2 marks)

Show Answers Only

a.    \(A=\{1,2,3,4,6,8,12,24\}\)

b.    \(A \cap B=\{1,4\}\)

c.    \((A \cup B) \cap C^c=\{1,2,4,8,16\}\)

Show Worked Solution

a.    \(A=\{1,2,3,4,6,8,12,24\}\)
 

b.    \(B=\{1,4,9,16\}\)

\(A \cap B=\{1,4\}\)
 

c.    \(A \cup B=\{1,2,3,4,6,8,9,12,16,24\}\)

\(C=\{3,6,9,12,15,18,21,24\}\)

\(C^c=\{1,2,4,5,7,8,10,11,13,14,16,17,19,20,22,23\}\)

\((A \cup B) \cap C^c=\{1,2,4,8,16\}\)

Probability, 2ADV EQ-Bank 12

Consider the universal set  \(U=\{x\) is a positive integer and \(x \leqslant 15\}\)

Three sets are defined as:

\begin{aligned}
& A=\{x \text { is a multiple of } 3\} \\
& B=\{x \text{ is a prime number}\} \\
& C=\{x \text{ is even}\}
\end{aligned}

  1. List the elements of set \(A\).   (1 mark)

  2.  Find  \(B \cap C\)   (1 mark)

  3. Find  \(A \cup \overline{C}\)   (2 marks)

Show Answers Only

a.    \(A=\{3,6,9,12,15\}\)

b.   \(B \cap C = \{2\} \)

c.    \(A \cup \overline{C}=\{1,3,5,6,7,9,11,12,13,15\}\)

Show Worked Solution

a.    \(A=\{3,6,9,12,15\}\)
 

b.    \(B=\{2,3,5,7,11,13\}\)

\(C=\{2,4,6,8,10,12,14\}\)

\(B \cap C = \{2\} \)
 

c.    \(\text{Find} \ \ A \cup \overline{C}:\)

\(\overline{C}=\{1,3,5,7,9,11,13\}\)

\(A \cup \overline{C}=\{1,3,5,6,7,9,11,12,13,15\}\)

Probability, 2ADV EQ-Bank 11

Consider the universal set  \(U=\{1,2,3,4,5,6,7,8,9,10\}\).

Two sets, \(A\) and \(B\), are given as

\(A= \{1,3,4,7,9\}\)

\(B =\{2,4,7,10\}\)

  1. Find \(A \cup B\)   (1 mark)

  2. Find \(A \cap \overline{B}\)   (2 marks)

Show Answers Only

a.    \(A \cup B = \{1,2,3,4,7,9,10\}\)

b.    \(A \cap \overline{B} = \{3, 4, 9\}\)

Show Worked Solution

a.    \(A= \{1,3,4,7,9\},\ \ B=\{2,4,7,10\}\)

\(A \cup B = \{1,2,3,4,7,9,10\}\)
 

b.     \(\overline{B} = \{1,3,5,6,8,9\}\)

\(A \cap \overline{B} = \{3, 4, 9\}\)

Trigonometry, 2ADV EQ-Bank 11

Prove the identity \(1+\tan ^2 \theta=\sec ^2 \theta\).   (2 marks)

Show Answers Only
\(\text{LHS}\) \(=1+\tan ^2 \theta\)
  \(=1+\dfrac{\sin ^2 \theta}{\cos ^2 \theta}\)
  \(=\dfrac{\cos ^2 \theta+\sin ^2 \theta}{\cos ^2 \theta}\)
  \(=\dfrac{1}{\cos ^2 \theta}\)
  \(=\sec ^2 \theta\)
Show Worked Solution

\(\text{Prove}\ \ 1+\tan ^2 \theta=\sec ^2 \theta\)

\(\text{LHS}\) \(=1+\tan ^2 \theta\)
  \(=1+\dfrac{\sin ^2 \theta}{\cos ^2 \theta}\)
  \(=\dfrac{\cos ^2 \theta+\sin ^2 \theta}{\cos ^2 \theta}\)
  \(=\dfrac{1}{\cos ^2 \theta}\)
  \(=\sec ^2 \theta\)

Calculus, 2ADV C1 EQ-Bank 14

A drone travels vertically from its launch pad.

It's height above ground, \(h\) metres, at time \(t\) minutes is modelled by

\(h(t)=-0.2 t^3+3 t^2+5 t\)  for  \(0 \leq t \leq 12\)

  1. Find the velocity of the drone at time \(t\) minutes.   (1 mark)

  2. Determine the exact time interval during which the drone is descending.   (2 marks)

Show Answers Only

a.    \(\dfrac{dh}{dt}=-0.6 t^2+6 t+5\)

b.    \(\dfrac{15+10 \sqrt{3}}{3}<t \leqslant 12\)

Show Worked Solution

a.    \(h=-0.2 t^3+3 t^2+5 t\)

\(\text{Velocity of the drone}=\dfrac{d h}{d t}.\)

\(\dfrac{dh}{dt}=-0.6 t^2+6 t+5\)
 

b.    \(\text{Drone is descending when} \ \ \dfrac{dh}{dt}<0:\)

\(-0.6 t^2+6 t+5\) \(<0\)  
\(0.6 t^2-6 t-5\) \(>0\)  
\(6 t^2-60 t-50\) \(>0\)  

 
\(\text{Solve}\ \ 6 t^2-60 t-50=0:\)

\(t=\dfrac{60 \pm \sqrt{(-60)^2+4 \times 6 \times 50}}{2 \times 6}=\dfrac{60 \pm \sqrt{4800}}{12}=\dfrac{15 \pm 10 \sqrt{3}}{3}\)

 
\(\text{Since parabola is concave up:}\)

\(6 t^2-60 t-50>0\ \ \text{when}\ \ t>\dfrac{15+10 \sqrt{3}}{3} \quad\left( t=\dfrac{15-10 \sqrt{3}}{3}<0\right)\)

\(\therefore \text{Drone is descending for} \ \ \dfrac{15+10 \sqrt{3}}{3}<t \leqslant 12\)

Calculus, 2ADV C1 EQ-Bank 18

The volume of water in a tank, \(V\) litres, at time \(t\) minutes is given by:

\(V(t)=2 t^3-15 t^2+24 t+50\)  for  \(0 \leqslant t \leqslant 6\)

  1. Find an expression for the rate at which water is flowing at time \(t\).   (1 mark)

  2. Calculate the rate of flow at  \(t=4\)  minutes and interpret this value.   (1 mark)

  3. Deduce when the water level in the tank is increasing.   (2 marks)

Show Answers Only

a.    \(\dfrac{dV}{d t}=6 t^2-30 t+24\)
 

b.    \(\text{At} \ \ t=4:\ \ \dfrac{dV}{d t}=0 \ \text{litres/minute}\)

\(\text{Interpretation: At \(\ t=4 \ \), water has stopped flowing either into or}\)

\(\text{out of the tank.}\)
 

c.    \(\text{If water level is increasing} \ \Rightarrow \ \dfrac{dV}{d t}>0:\)

\(\text {Find \(t\) when}\ \dfrac{dV}{d t}>0:\)

\(6 t^2-30t+24\) \(\gt 0\)  
\(6\left(t^2-5 t+4\right)\) \(\gt 0\)  
\((t-4)(t-1)\) \(\gt 0\)  

 

\(\therefore \ \text{Water level increases for}\ \ t \in [0,1) \cup (4, 6]\)

Show Worked Solution

a.    \(V=2 t^3-15 t^2+24 t+50\)

\(\dfrac{dV}{d t}=6 t^2-30 t+24\)
 

b.    \(\text{At} \ \ t=4:\)

\(\dfrac{dV}{d t}=6 \times 4^2-30 \times 4+24=0 \ \text{litres/minute}\)

\(\text{Interpretation: At \(\ t=4 \ \), water has stopped flowing either into or}\)

\(\text{out of the tank.}\)
 

c.    \(\text{If water level is increasing} \ \Rightarrow \ \dfrac{dV}{d t}>0:\)

\(\text {Find \(t\) when}\ \dfrac{dV}{d t}>0:\)

\(6 t^2-30t+24\) \(\gt 0\)  
\(6\left(t^2-5 t+4\right)\) \(\gt 0\)  
\((t-4)(t-1)\) \(\gt 0\)  

 

\(\therefore \ \text{Water level increases for}\ \ t \in [0,1) \cup (4, 6]\)

Functions, 2ADV EQ-Bank 21

Consider the function  \(g(x)=-\dfrac{12}{x}\)

  1. Is \(g(x)\) an odd or even function? Give reason(s) for your answer.   (2 marks)

  2. Sketch the graph \(y=g(x)\) in the domain \(-6 \leq x \leq 6\). Label the endpoints and two other points on the curve.   (2 marks)

Show Answers Only

a.    \(\text{If}\ g(x)\ \text{is odd}\ \ \Rightarrow \ \ g(-x)=-g(x)\)

\(-g(x)= \dfrac{12}{x}\)

\(g(-x) = -\dfrac{12}{(-x)} = \dfrac{12}{x} = -g(x)\)

\(\therefore g(x)\ \text{is odd (and cannot be even).}\)
 

b.    \(\text{Table of values:}\)

\begin{array}{|c|c|c|c|c|c|}
\hline \rule{0pt}{2.5ex}\ \ x \ \ \rule[-1ex]{0pt}{0pt}& -4 & -2 & \ \ 0 \ \ & 2 & 4 \\
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& 3 & 6 & \infty & -6 & -3 \\
\hline
\end{array}

\(\text{Endpoints:}\ (-6,2), (6,-2) \)
 

Show Worked Solution

a.    \(\text{If}\ g(x)\ \text{is odd}\ \ \Rightarrow \ \ g(-x)=-g(x)\)

\(-g(x)= \dfrac{12}{x}\)

\(g(-x) = -\dfrac{12}{(-x)} = \dfrac{12}{x} = -g(x)\)

\(\therefore g(x)\ \text{is odd.}\)
 

b.    \(\text{Table of values:}\)

\begin{array}{|c|c|c|c|c|c|}
\hline \rule{0pt}{2.5ex}\ \ x \ \ \rule[-1ex]{0pt}{0pt}& -4 & -2 & \ \ 0 \ \ & 2 & 4 \\
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& 3 & 6 & \infty & -6 & -3 \\
\hline
\end{array}

\(\text{Endpoints:}\ (-6,2), (6,-2) \)