A forest ranger wishes to investigate the mass of adult male koalas in a Victorian forest. A random sample of 20 such koalas has a sample mean of 11.39 kg. It is known that the mass of adult male koalas in the forest is normally distributed with a standard deviation of 1 kg. --- 3 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- The ranger wants to decrease the width of the 95% confidence interval by 60% to get a better estimate of the population mean. --- 2 WORK AREA LINES (style=lined) --- It is thought that the mean mass of adult male koalas in the forest is 12 kg. The ranger thinks that the true mean mass is less than this and decides to apply a one-tailed statistical test. A random sample of 40 adult male koalas is taken and the sample mean is found to be 11.6 kg. --- 2 WORK AREA LINES (style=lined) --- The ranger decides to apply the one-tailed test at the 1% level of significance and assumes the mass of adult male koalas in the forest is normally distributed with a mean of 12 kg and a standard deviation of 1 kg. --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- Suppose that the true mean mass of adult male koalas in the forest is 11.4 kg, and the standard deviation is 1 kg. The level of significance of the test is still 1%. --- 2 WORK AREA LINES (style=lined) ---
Functions, MET2 2023 VCAA 2
The following diagram represents an observation wheel, with its centre at point \(P\). Passengers are seated in pods, which are carried around as the wheel turns. The wheel moves anticlockwise with constant speed and completes one full rotation every 30 minutes.When a pod is at the lowest point of the wheel (point \(A\)), it is 15 metres above the ground. The wheel has a radius of 60 metres.
Consider the function \(h(t)=-60\ \cos(bt)+c\) for some \(b, c \in R\), which models the height above the ground of a pod originally situated at point \(A\), after time \(t\) minutes.
- Show that \(b=\dfrac{\pi}{15}\) and \(c=75\). (2 marks)
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- Find the average height of a pod on the wheel as it travels from point \(A\) to point \(B\).
- Give your answer in metres, correct to two decimal places. (2 marks)
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- Find the average rate of change, in metres per minute, of the height of a pod on the wheel as it travels from point \(A\) to point \(B\). (1 mark)
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After 15 minutes, the wheel stops moving and remains stationary for 5 minutes. After this, it continues moving at double its previous speed for another 7.5 minutes.
The height above the ground of a pod that was initially at point \(A\), after \(t\) minutes, can be modelled by the piecewise function \(w\):
\(w(t) = \begin {cases}
h(t) &\ \ 0 \leq t < 15 \\
k &\ \ 15 \leq t < 20 \\
h(mt+n) &\ \ 20\leq t\leq 27.5
\end{cases}\)
where \(k\geq 0, m\geq 0\) and \(n \in R\).
-
- State the values of \(k\) and \(m\). (1 mark)
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- State the values of \(k\) and \(m\). (1 mark)
-
- Find all possible values of \(n\). (2 marks)
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- Sketch the graph of the piecewise function \(w\) on the axes below, showing the coordinates of the endpoints. (3 marks)
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- Find all possible values of \(n\). (2 marks)
Show Answers Only
a. \(\text{See worked solution}\)
b. \(\approx 36.80\ \text{m}\)
c. \(8\)
d.i. \(k=135, m=2\)
d.ii. \(n=30p+5,\ p\in Z\)
d.iii.
Show Worked Solution
| a. | \(\text{Period:}\) | \(\dfrac{2\pi}{b}\) | \(=30\) |
| \(\therefore\ b\) | \(=\dfrac{\pi}{15}\) |
\(\text{Given }h(t)=-60\cos(bt)+c,\ \text{evaluate when }t=0, h=15\ \text{to find }c.\)
| \(-60\ \cos(0)+c\) | \(=15\) |
| \(c\) | \(=75\) |
\(\therefore\ h(t)=-60\cos(\dfrac{\pi t}{15})+75\)
| b. | \(\text{Average height}\) | \(=\dfrac{1}{\frac{30}{4}-0}\displaystyle\int_0^{\frac{30}{4}}\Bigg(-60\cos\Bigg(\dfrac{\pi}{15}t\Bigg)+75\Bigg)\,dt\) |
| \(=\dfrac{2}{15}\left[75t-\dfrac{900\ \sin(\frac{\pi}{15}t)}{\pi}\right]_{0}^{7.5}\) | ||
| \(=\dfrac{2}{15}\Bigg[75\times 7.5-\dfrac{900\ \sin(\frac{\pi}{15}\times 7.5)}{\pi}\Bigg]-\Bigg[0\Bigg]\) | ||
| \(=\dfrac{75\pi-120}{\pi}=\dfrac{15(5\pi-8)}{\pi}\) | ||
| \(=36.802\dots\approx 36.80\ \text{m}\) |
♦♦ Mean mark (b) 45%.
MARKER’S COMMENT: \(\frac{1}{60}\int_0^{60} h(t)dt\) was a common error.
Others incorrectly found average rate of change instead of average value.
MARKER’S COMMENT: \(\frac{1}{60}\int_0^{60} h(t)dt\) was a common error.
Others incorrectly found average rate of change instead of average value.
| c. | \(\text{Av rate of change of height}\) | \(=\dfrac{h(7.5)-h(0)}{7.5}\) |
| \(=\dfrac{\Bigg(75-60\cos(\dfrac{\pi \times 7.5}{15})\Bigg)-\Bigg(75-60\cos(\dfrac{\pi\times 0}{15})\Bigg)}{7.5}\) | ||
| \(=\dfrac{75-15}{7.5}=8\) |
♦ Mean mark (c) 50%.
MARKER’S COMMENT: Common incorrect answer was – 8.
MARKER’S COMMENT: Common incorrect answer was – 8.
| di. | \(\text{Period is 30 minutes, so after 15 minutes pod}\) \(\text{is at the top of the wheel.}\) |
| \(\therefore\ k=75-60\ \cos(\frac{\pi}{15}\times 15)=135\) |
\(\text{Pod is travelling at twice its previous speed}\)
\(\text{so one revolution takes 15 minutes}\)
| \(\therefore\ \text{Period}\rightarrow\) | \(\dfrac{2\pi}{bm}\) | \(=15\) |
| \(bm\) | \(=\dfrac{2\pi}{15}\) | |
| \(\dfrac{\pi}{15}\cdot m\) | \(=\dfrac{2\pi}{15}\) | |
| \(m\) | \(=\dfrac{2\pi}{15}\times \dfrac{15}{\pi}\) | |
| \(=2\) |
♦♦ Mean mark (d)(i) 40%.
MARKER’S COMMENT: Many students could find \(k\) but not \(m\) with \(m=\frac{1}{2}\) a common error.
MARKER’S COMMENT: Many students could find \(k\) but not \(m\) with \(m=\frac{1}{2}\) a common error.
| dii. | \(\text{When }t=20\ \text{the pod is at the top of the wheel and height is 135.}\) \(\text{When }t=27.5\ \text{the pod is back at the start and height is 15.}\) |
\(\text{Using CAS solve for }t=20:\rightarrow\ h(2(20)+n)=135\ \rightarrow\ n=30p+5\)
\(\text{Using CAS solve for }t=27.5:\rightarrow\ h(2(27.5)+n)=15\ \rightarrow\ n=30p+5\)
\(\therefore\ n=30p+5,\ p\in Z\)
♦♦♦ Mean mark (d)(ii) 20%.
MARKER’S COMMENT: Many students set up the correct equations to solve but did not provide the general solution. Some incorrectly stated the variable as an element of R.
MARKER’S COMMENT: Many students set up the correct equations to solve but did not provide the general solution. Some incorrectly stated the variable as an element of R.
diii.
♦♦♦ Mean mark (d)(iii) 40%.
MARKER’S COMMENT: Students are reminded to include all endpoints and be particular about the curvature of graph.
MARKER’S COMMENT: Students are reminded to include all endpoints and be particular about the curvature of graph.
PHYSICS, M7 2021 VCE 20 MC
One of Einstein's postulates for special relativity is that the laws of physics are the same in all inertial frames of reference.
Which one of the following best describes a property of an inertial frame of reference?
- It is travelling at a constant speed.
- It is travelling at a speed much slower than \(c\).
- Its movement is consistent with the expansion of the universe.
- No observer in the frame can detect any acceleration of the frame.
PHYSICS, M8 2021 VCE 18*
A monochromatic light source is emitting green light with a wavelength of 550 nm. The light source emits 2.8 × 10\(^{16}\) photons every second.
Calculate the power of the light source in Watts? (2 marks)
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PHYSICS, M7 2021 VCE 16 MC
The diagram below shows a circuit that is used to study the photoelectric effect.
Which one of the following is essential to the measurement of the maximum kinetic energy of the emitted photoelectrons?
- the level of brightness of the light source
- the wavelengths that pass through the filter
- the reading on the voltmeter when the current is at a minimum value
- the reading on the ammeter when the voltage is at a maximum value
PHYSICS, M6 2021 VCE 5 MC
The diagram below shows a small DC electric motor, powered by a battery that is connected via a split-ring commutator. The rectangular coil has sides KJ and LM. The magnetic field between the poles of the magnet is uniform and constant.
The switch is now closed, and the coil is stationary and in the position shown in the diagram.
Which one of the following statements best describes the motion of the coil when the switch is closed?
- The coil will remain stationary.
- The coil will rotate in direction A, as shown in the diagram.
- The coil will rotate in direction B, as shown in the diagram.
- The coil will oscillate regularly between directions A and B, as shown in the diagram.
PHYSICS, M5 2021 VCE 4*
The planet Phobetor has a mass four times that of Earth. Acceleration due to gravity on the surface of Phobetor is 18 m s\(^{-2}\).
If Earth has a radius \(R\), calculate the radius of Phobetor in terms of \(R\)? (2 marks)
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PHYSICS, M8 2022 VCE 17
A materials scientist is studying the diffraction of electrons through a thin metal foil. She uses electrons with an energy of 10.0 keV. The resulting diffraction pattern is shown in Figure 19. --- 6 WORK AREA LINES (style=lined) --- The materials scientist then increases the energy of the electrons by a small amount and hence their speed by a small amount. Explain what effect this would have on the de Broglie wavelength of the electrons. Justify your answer. (3 marks) --- 6 WORK AREA LINES (style=lined) ---
Functions, MET2 2023 VCAA 20 MC
Let \(f(x)=\log_{e}\Bigg(x+\dfrac{1}{\sqrt{2}}\Bigg)\).
Let \(g(x)=\sin(x)\) where \(x\in (-\infty, 5)\).
The largest interval of \(x\) values for which \((f\circ g)(x)\) and \((g\circ f)(x)\) both exist is
- \(\Bigg(-\dfrac{1}{\sqrt{2}},\dfrac{5\pi}{4}\Bigg)\)
- \(\Bigg[-\dfrac{1}{\sqrt{2}},\dfrac{5\pi}{4}\Bigg)\)
- \(\Bigg(-\dfrac{\pi}{4},\dfrac{5\pi}{4}\Bigg)\)
- \(\Bigg[-\dfrac{\pi}{4},\dfrac{5\pi}{4}\Bigg]\)
- \(\Bigg[-\dfrac{\pi}{4},-\dfrac{1}{\sqrt{2}}\Bigg]\)
Functions, 2ADV F2 2023 MET1 3
- Sketch the graph of \(f(x)=2-\dfrac{3}{x-1}\) on the axes below, labelling all asymptotes with their equation and axial intercepts with their coordinates. (3 marks)
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- Find the values of \(x\) for which \(f(x)\leq1\). (1 mark)
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Show Answers Only
a.
b. \(1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]\)
Show Worked Solution
a. \(\text{Vertical asymptote when}\ \ x=1\)
\(y\text{-int:}\ y=2-(-3)\ \ \Rightarrow\ \ y=5\)
\(x\text{-int:}\ 2-\dfrac{3}{x-1}=0\ \ \Rightarrow\ \ x=\dfrac{5}{2} \)
\(\text{As}\ \ x \rightarrow \infty, \ \ y \rightarrow 2^{-}; \ \ x \rightarrow -\infty, \ \ y \rightarrow 2^{+} \)
b. \(\text{From the graph:}\)
\(f(x)=1\ \text{when }x=4\)
\(x>1\ \text{to the right of the vertical asymptote}\)
\(\therefore\ f(x)\leq1\ \text{when}\ \ 1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]\)
♦♦ Mean mark (b) 38%.
MARKER’S COMMENT: Many students did not use their graph from part (a).
MARKER’S COMMENT: Many students did not use their graph from part (a).
PHYSICS, M7 2022 VCE 11
Explain why muons formed in the outer atmosphere can reach the surface of Earth even though their half-lives indicate that they should decay well before reaching Earth's surface. (2 marks)
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PHYSICS, M5 2022 VCE 8
A Formula 1 racing car is travelling at a constant speed of 144 km h\(^{-1}\) (40 m s\(^{-1}\)) around a horizontal corner of radius 80.0 m. The combined mass of the driver and the car is 800 kg. Figure 8a shows a front view and Figure 8b shows a top view. --- 4 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=blank) --- --- 5 WORK AREA LINES (style=lined) --- a. \(1.6 \times 10^4\ \text{N}\) b. c. → A net horizontal force is required for circular motion. → The horizontal force is perpendicular to the direction of travel. → This is the force which changes the direction of travel. → The force of friction from the road acts on the tires provides the centripetal force in this motion. a. \(F=\dfrac{mv^2}{r}=\dfrac{800 \times (40)^2}{80}=1.6 \times 10^4\ \text{N}\) b. c. → A net horizontal force is required for circular motion. → The horizontal force is perpendicular to the direction of travel. → This is the force which changes the direction of travel. → The force of friction from the road acts on the tires provides the centripetal force in this motion.
Show Answers Only
Show Worked Solution
♦ Mean mark 42%.
PHYSICS, M8 2022 VCE 9
A star is transforming energy at a rate of 2.90 × 10\(^{25}\) W.
Explain the type of transformation involved and what effect, if any, the transformation would have on the mass of the star. No calculations are required. (2 marks)
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PHYSICS, M6 2022 VCE 4
A square loop of wire connected to a resistor, \(\text{R}\), is placed close to a long wire carrying a constant current, \(I\), in the direction shown in Figure 4.
The square loop is moved three times in the following order:
- Movement A – Starting at Position 1 in Figure 4, the square loop rotates one full rotation at a steady speed about the \(x\)-axis. The rotation causes the resistor, \(\text{R}\), to first move out of the page.
- Movement B – The square loop is then moved at a constant speed, parallel to the current carrying wire, from Position 1 to Position 2 in Figure 4.
- Movement C – The square loop is moved at a constant speed, perpendicular to the current carrying wire, from Position 2 to Position 3 in Figure 4.
Complete the table below to show the effects of each of the three movements by:
- sketching any EMF generated in the square loop during the motion on the axes provided (scales and values are not required)
- stating whether any induced current in the square loop is 'alternating', 'clockwise', 'anticlockwise' or has 'no current'.
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Show Answers Only
Show Worked Solution
The magnetic field strength around a current carrying conductor, \(B \propto \dfrac{1}{r}\). Hence, for the movement in C, the change in magnetic field strength and therefore magnetic flux is one of inverse proportionality as depicted in the graph below.
♦ Mean mark 43%.
Calculus, SPEC2 2023 VCAA 4
A fish farmer releases 200 fish into a pond that originally contained no fish. The fish population, \(P\), grows according to the logistic model, \(\dfrac{d P}{d t}=P\left(1-\dfrac{P}{1000}\right)\) , where \(t\) is the time in years after the release of the 200 fish. --- 2 WORK AREA LINES (style=lined) --- One form of the solution for \(P\) is \(P=\dfrac{1000}{1+D e^{-t}}\ \), where \(D\) is a real constant. --- 2 WORK AREA LINES (style=lined) --- The farmer releases a batch of \(n\) fish into a second pond, pond 2 , which originally contained no fish. The population, \(Q\), of fish in pond 2 can be modelled by \(Q=\dfrac{1000}{1+9 e^{-1.1 t}}\), where \(t\) is the time in years after the \(n\) fish are released. --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) --- The farmer wishes to take 5.5% of the fish from pond 2 each year. The modified logistic differential equation that would model the fish population, \(Q\), in pond 2 after \(t\) years in this situation is \(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)-0.055Q\) --- 4 WORK AREA LINES (style=lined) --- a. \(A=1, \quad B=\dfrac{1}{1000}\) b. \(\Rightarrow D=4\) c. \(n=\dfrac{1000}{1+9 e^0}=100\) d. \(Q=\dfrac{1000}{1+9 e^{-6.6}} \approx 988\) e.i. \(Q^{\prime}=\frac{121}{100}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{121}{100\ 000}\left(1-\frac{Q}{1000}\right)\) e.ii. \(t=2\) f. g. \(Q=950\) a. \(\dfrac{dP}{dt}=P\left(1-\dfrac{P}{1000}\right) \ \Rightarrow \ \dfrac{d t}{d P}=\dfrac{1}{P}\left(\dfrac{1}{1-\frac{P}{1000}}\right)\) \(\text{Expand (by CAS):}\) \(\dfrac{1}{P}\left(\dfrac{1}{1-\frac{P}{1000}}\right)=\dfrac{1}{P}-\dfrac{1}{(P-1000)}=\dfrac{1}{P}+\dfrac{1}{1000}\left(\dfrac{1}{1-\frac{P}{1000}}\right)\) \(A=1, \quad B=\dfrac{1}{1000}\) b. \(\text{When}\ \ t=0, P=200 \text{ (given)}\) \(\text{Solve}\ \ 200=\dfrac{1000}{1+D e^0}\ \ \text{for}\ t:\) \(\Rightarrow D=4\ \ \text {(by CAS):}\) c. \(Q=\dfrac{1000}{1+9 e^{-1.1 t}}\) \(\text{At}\ \ t=0, Q=n\): \(n=\dfrac{1000}{1+9 e^0}=100\) d. \(\text{Find } Q \text{ when } t=6:\) \(Q=\dfrac{1000}{1+9 e^{-6.6}} \approx 988\) e.i. \(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)\) \(\text{Using the product rule:}\) \begin{aligned} e.ii. \(\text{Max } Q^{\prime} \Rightarrow Q^{\prime \prime}=0\) \(\text{Solve } Q^{\prime \prime}=0 \ \ \text {(by CAS):}\) \(\Rightarrow Q=500\) \(\text{Solve } Q=500 \text{ for } t \text{ (by CAS):}\) \(t=1.99 \ldots=2 \ \text{(nearest year)}\) f. g. \(\text{Solve for } Q \ \text{(by CAS):}\) \(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)-0.055\,Q=0\) \(\Rightarrow Q=950\)
Show Answers Only
Show Worked Solution
♦ Mean mark (a) 48%.
Q^{\prime \prime} & =\frac{11}{10}\left[Q^{\prime}\left(1-\frac{Q}{1000}\right)+Q\left(-\frac{1}{1000}\, Q^{\prime}\right)\right] \\
& =\frac{11}{10}\left[\frac{11}{10}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{Q}{1000}\left(\frac{11}{10}\, Q\left(1-\frac{Q}{1000}\right)\right)\right] \\
& =\frac{121}{100}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{121}{100\ 000}\left(1-\frac{Q}{1000}\right)
\end{aligned}
♦♦♦ Mean mark (e)(i) 21%.
♦ Mean mark (g) 40%.
PHYSICS, M7 2023 VCE 13
A group of physics students undertake a Young's double-slit experiment using the apparatus shown in Figure 15. They use a green laser that produces light with a wavelength of 510 nm. The light is incident on two narrow slits, S\(_1\) and S\(_2\). The distance between the two slits is 100 \( \mu \)m.
An interference pattern is observed on a screen with points P\(_{0}\), P\(_{1}\) and P\(_2\) being the locations of adjacent bright bands, shown in Figure 15. Point P\(_0\) is the central bright band.
- Calculate the path difference between S\(_{1}\)P\(_{2}\) and S\(_{2}\)P\(_{2}\). Give your answer in metres. Show your working. (2 marks)
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- The green laser is replaced by a red laser.
- Describe the effect of this change on the spacing between adjacent bright bands. (1 mark)
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- Explain how Young's double-slit experiment provides evidence for the wave-like nature of light and not the particle-like nature of light. (3 marks)
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PHYSICS, M5 2023 VCE 9
Giorgos is practising his tennis serve using a tennis ball of mass 56 g. --- 5 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) --- --- 7 WORK AREA LINES (style=lined) ---
Calculus, SPEC2 2023 VCAA 3
The curve given by \(y^2=x-1\), where \(2 \leq x \leq 5\), is rotated about the \(x\)-axis to form a solid of revolution. --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- The total surface area of the solid consists of the curved surface area plus the areas of the two circular discs at each end. The 'efficiency ratio' of a body is defined as its total surface area divided by the enclosed volume. --- 2 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) ---
Functions, MET2 2023 VCAA 19 MC
Find all the values of \(k\), such that the equation \(x^2+(4k+3)x+4k^2-\dfrac{9}{4}=0\) has two real solutions for \(x\), one positive and one negative.
- \(k>-\dfrac{3}{4}\)
- \(k\geq-\dfrac{3}{4}\)
- \(k>\dfrac{3}{4}\)
- \(-\dfrac{3}{4}<k<\dfrac{3}{4}\)
- \(k<-\dfrac{3}{4}\ \text{or}\ k>\dfrac{3}{4}\)
Graphs, MET2 2023 VCAA 18 MC
Consider the function \(f:[-a\pi, a\pi] \rightarrow\ R, f(x)=\sin(ax)\), where \(a\) is a positive integer.
The number of local minima in the graph of \(y=f(x)\) is always equal to
- \(2\)
- \(4\)
- \(a\)
- \(2a\)
- \(a^2\)
Show Worked Solution
\(\text{Check graphs for different values of }a\)
\(a=1\to \text{1 local minimum}\)
\(a=2\to \text{4 local minimums}\)
\(a=3\to \text{9 local minimums}\)
\(\therefore\ \text{Number of local minimums is always equal to }a^2\)
\(\Rightarrow E\)
♦♦ Mean mark 29%.
MARKER’S COMMENT: 22% incorrectly chose C and 26% incorrectly chose D.
MARKER’S COMMENT: 22% incorrectly chose C and 26% incorrectly chose D.
Calculus, MET2 2023 VCAA 17 MC
A cylinder of height \(h\) and radius \(r\) is formed from a thin rectangular sheet of metal of length \(x\) and \(y\), by cutting along the dashed lines shown below.
The volume of the cylinder, in terms of \(x\) and \(y\), is given by
- \(\pi x^2y\)
- \(\dfrac{\pi xy^2-2y^3}{4\pi^2}\)
- \(\dfrac{2y^3-\pi xy^2}{4\pi^2}\)
- \(\dfrac{\pi xy-2y^2}{2\pi}\)
- \(\dfrac{2y^2-\pi xy}{2\pi}\)
Probability, MET2 2023 VCAA 4
A manufacturer produces tennis balls.
The diameter of the tennis balls is a normally distributed random variable \(D\), which has a mean of 6.7 cm and a standard deviation of 0.1 cm.
- Find \(\Pr(D>6.8)\), correct to four decimal places. (1 mark)
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- Find the minimum diameter of a tennis ball that is larger than 90% of all tennis balls produced.
Give your answer in centimetres, correct to two decimal places. (1 mark)--- 2 WORK AREA LINES (style=lined) ---
Tennis balls are packed and sold in cylindrical containers. A tennis ball can fit through the opening at the top of the container if its diameter is smaller than 6.95 cm.
- Find the probability that a randomly selected tennis ball can fit through the opening at the top of the container.
Give your answer correct to four decimal places. (1 mark)--- 2 WORK AREA LINES (style=lined) ---
- In a random selection of 4 tennis balls, find the probability that at least 3 balls can fit through the opening at the top of the container.
Give your answer correct to four decimal places. (2 marks)--- 2 WORK AREA LINES (style=lined) ---
A tennis ball is classed as grade A if its diameter is between 6.54 cm and 6.86 cm, otherwise it is classed as grade B.
- Given that a tennis ball can fit through the opening at the top of the container, find the probability that it is classed as grade A.
Give your answer correct to four decimal places. (2 marks)--- 4 WORK AREA LINES (style=lined) ---
- The manufacturer would like to improve processes to ensure that more than 99% of all tennis balls produced are classed as grade A.
Assuming that the mean diameter of the tennis balls remains the same, find the required standard deviation of the diameter, in centimetres, correct to two decimal places. (2 marks)--- 5 WORK AREA LINES (style=lined) ---
- An inspector takes a random sample of 32 tennis balls from the manufacturer and determines a confidence interval for the population proportion of grade A balls produced.
The confidence interval is (0.7382, 0.9493), correct to four decimal places.
Find the level of confidence that the population proportion of grade A balls is within the interval, as a percentage correct to the nearest integer. (2 marks)--- 5 WORK AREA LINES (style=lined) ---
A tennis coach uses both grade A and grade B balls. The serving speed, in metres per second, of a grade A ball is a continuous random variable, \(V\), with the probability density function
\(f(v) = \begin {cases}
\dfrac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg) &\ \ 30 \leq v \leq 3\pi^2+30 \\
0 &\ \ \text{elsewhere}
\end{cases}\)
- Find the probability that the serving speed of a grade A ball exceeds 50 metres per second.
Give your answer correct to four decimal places. (1 mark)--- 2 WORK AREA LINES (style=lined) ---
- Find the exact mean serving speed for grade A balls, in metres per second. (1 mark)
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The serving speed of a grade B ball is given by a continuous random variable, \(W\), with the probability density function \(g(w)\).
A transformation maps the graph of \(f\) to the graph of \(g\), where \(g(w)=af\Bigg(\dfrac{w}{b}\Bigg)\).
- If the mean serving speed for a grade B ball is \(2\pi^2+8\) metres per second, find the values of \(a\) and \(b\). (2 marks)
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PHYSICS, M6 2023 VCE 6
Kim and Charlie are attempting to create a DC generator and have arranged the magnets along the axis of rotation of the wire loop, J, K, L and M, as shown in Figure 6. They are having some trouble getting it to work. They rotate the loop in the direction of the arrow, as shown in Figure 6. --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- a. → The generator has the magnetic field parallel to the plane of the loop. → Therefore, there is no flux cutting through the loop of wire. → Further, as the loop rotates there is no change in flux and no current or EMF produced. b. → The magnets should be rotated by 90°. a. → The generator has the magnetic field parallel to the plane of the loop. → Therefore, there is no flux cutting through the loop of wire. → Further, as the loop rotates there is no change in flux and no current or EMF produced. b. → The magnets should be rotated by 90°.
Show Answers Only
Show Worked Solution
♦ Mean mark (a) 46%.
PHYSICS, M6 2023 VCE 5
Figure 4a shows a single square loop of conducting wire placed just outside a constant uniform magnetic field, \(B\). The length of each side of the loop is 0.040 m. The magnetic field has a magnitude of 0.30 T and is directed out of the page.
Over a time period of 0.50 s, the loop is moved at a constant speed, \(v\), from completely outside the magnetic field, Figure 4a, to completely inside the magnetic field, Figure 4b.
- Calculate the average EMF produced in the loop as it moves from the position just outside the region of the field, Figure \(4 \mathrm{a}\), to the position completely within the area of the magnetic field, Figure 4b.
- Show your working. (2 marks)
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- On the small square loop in Figure 5, show the direction of the induced current as the loop moves into the area of the magnetic field. (1 mark)
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Show Answers Only
a. \(9.6 \times 10^{-4}\ \text{V}\)
b.
Show Worked Solution
| a. | \(\varepsilon\) | \(=\dfrac{\Delta \phi}{\Delta t}\) |
| \(=\dfrac{\Delta B A}{\Delta t}\) | ||
| \(=\dfrac{0.3 \times 0.04^2}{0.5}\) | ||
| \(=9.6 \times 10^{-4}\ \text{V}\) |
b.
♦ Mean mark (b) 45%.
Calculus, MET2 2023 VCAA 5
Let \(f:R \to R, f(x)=e^x+e^{-x}\) and \(g:R \to R, g(x)=\dfrac{1}{2}f(2-x)\).
- Complete a possible sequence of transformations to map \(f\) to \(g\). (2 marks)
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Two functions \(g_1\) and \(g_2\) are created, both with the same rule as \(g\) but with distinct domains, such that \(g_1\) is strictly increasing and \(g_2\) is strictly decreasing.
- Give the domain and range for the inverse of \(g_1\). (2 marks)
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Shown below is the graph of \(g\), the inverse of \(g_1\) and \(g_2\), and the line \(y=x\).
The intersection points between the graphs of \(y=x, y=g(x)\) and the inverses of \(g_1\) and \(g_2\), are labelled \(P\) and \(Q\).
-
- Find the coordinates of \(P\) and \(Q\), correct to two decimal places. (1 mark)
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- Find the coordinates of \(P\) and \(Q\), correct to two decimal places. (1 mark)
-
- Find the area of the region bound by the graphs of \(g\), the inverse of \(g_1\) and the inverse of \(g_2\).
Give your answer correct to two decimal places. (2 marks)
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- Find the area of the region bound by the graphs of \(g\), the inverse of \(g_1\) and the inverse of \(g_2\).
Let \(h:R\to R, h(x)=\dfrac{1}{k}f(k-x)\), where \(k\in (o, \infty)\).
- The turning point of \(h\) always lies on the graph of the function \(y=2x^n\), where \(n\) is an integer.
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Let \(h_1:[k, \infty)\to R, h_1(x)=h(x)\).
The rule for the inverse of \(h_1\) is \(y=\log_{e}\Bigg(\dfrac{1}{k}x+\dfrac{1}{2}\sqrt{k^2x^2-4}\Bigg)+k\)
- What is the smallest value of \(k\) such that \(h\) will intersect with the inverse of \(h_1\)?
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It is possible for the graphs of \(h\) and the inverse of \(h_1\) to intersect twice. This occurs when \(k=5\).
- Find the area of the region bound by the graphs of \(h\) and the inverse of \(h_1\), where \(k=5\).
Give your answer correct to two decimal places. (2 marks)
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PHYSICS, M6 2023 VCE 4*
A transformer is used to provide a low-voltage supply for six outdoor garden globes. The circuit is shown in Figure 3. Assume there is no power loss in the connecting wires. The input of the transformer is connected to a power supply that provides an AC voltage of 240 V. The globes in the circuit are designed to operate with an AC voltage of 12 V. Each globe is designed to operate with a power of 20 W. --- 2 WORK AREA LINES (style=lined) --- The globes are turned on. --- 4 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) ---
PHYSICS, M5 2023 VCE 2
Phobos is a small moon in a circular orbit around Mars at an altitude of 6000 km above the surface of Mars. The gravitational field strength of Mars at its surface is 3.72 N kg\(^{-1}\). The radius of Mars is 3390 km. --- 5 WORK AREA LINES (style=lined) --- --- 7 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) ---
PHYSICS, M5 2023 VCE 3 MC
Space scientists want to place a satellite into a circular orbit where the gravitational field strength of Earth is half of its value at Earth's surface.
Which one of the following expressions best represents the altitude of this orbit above Earth's surface, where \(R\) is the radius of Earth?
- \(\dfrac{\sqrt{2} R}{2}-R\)
- \(\sqrt{2} R\)
- \((\sqrt{2} R)-R\)
- \(2 R-\sqrt{2} R\)
Statistics, SPEC2 2023 VCAA 20 MC
The lifespan of a certain electronic component is normally distributed with a mean of \(\mu\) hours and a standard deviation of \(\sigma\) hours.
Given that a 99% confidence interval, based on a random sample of 100 such components, is (10 500, 15 500), the value of \(\sigma\) is closest to
- 9710
- 10 750
- 12 750
- 15 190
- 19 390
Calculus, MET2 2023 VCAA 14 MC
A polynomial has the equation \(y=x(3x-1)(x+3)(x+1)\).
The number of tangents to this curve that pass through the positive \(x\)-intercept is
- 0
- 1
- 2
- 3
- 4
Vectors, EXT1 V1 SM-Bank 31
The sum of two unit vectors is a unit vector.
Determine the magnitude of the difference of the two vectors. (3 marks)
Show Worked Solution
\(\text{Vectors can be drawn as two sides of an equilateral triangle.}\)
\(\text{Using the cosine rule for the difference between the two vectors:}\)
| \(c^2\) | \(=a^2+b^2-2ab\, \cos C\) | |
| \(=1+1-2\times -\dfrac{1}{2} \) | ||
| \(=3\) | ||
| \(c\) | \(=\sqrt{3}\) |
\(\abs{v_1-v_2} = \sqrt{3}\)
Vectors, SPEC2 2023 VCAA 16 MC
A student throws a ball for his dog to retrieve. The position vector of the ball, relative to an origin \(O\) at ground level \(t\) seconds after release, is given by \( \underset{\sim}{\text{r}}{}_\text{B} (t)=5 t \underset{\sim}{\text{i}}+7 t \underset{\sim}{\text{j}}+(15 t-4.9 t^2+1.5) \underset{\sim}{\text{k}} \). Displacement components are measured in metres, where \(\underset{\sim}{\text{i}}\) is a unit vector to the east, \(\underset{\sim}{\text{j}}\) is a unit vector to the north and \( \underset{\sim} {\text{k}}\) is a unit vector vertically up.
The total \( \textbf{vertical} \) distance, in metres, travelled by the ball before it hits the ground is closest to
- 1.5
- 11.5
- 13.0
- 24.5
- 26.0
Vectors, SPEC2 2023 VCAA 14 MC
Let \(\underset{\sim}{\text{a}}=\underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}, \underset{\sim}{\text{b}}=\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}\) and \(\text{c}=\underset{\sim}{\text{i}}+2 \underset{\sim}{\text{j}}+3 \underset{\sim}{\text{k}}\).
If \(\underset{\sim}{\text{n}}\) is a unit vector such that \( \underset{\sim} {\text{a}} \cdot \underset{\sim} {\text{n}}=0\) and \( \underset{\sim} {\text{b}} \cdot \underset{\sim} {\text{n}}=0\), then \(\big{|} \underset{\sim} {\text{c}} \cdot \underset{\sim} {\text{n}}\big{|} \) is equal to
- 2
- 3
- 4
- 5
- 6
Calculus, SPEC2 2023 VCAA 12 MC
The acceleration, \(a\) ms\(^{-2}\), of a particle that starts from rest and moves in a straight line is described by \(a=1+v\), where \(v\) ms\(^{-1}\) is its velocity after \(t\) seconds.
The velocity of the particle after \( \log _e(e+1) \) seconds is
- \(e\)
- \(e+1\)
- \(e^2+1\)
- \(\log _e(1+e)+1\)
- \(\log _e\left(\log _e(1+e)-1\right)\)
Calculus, MET2 2023 VCAA 13 MC
The following algorithm applies Newton's method using a For loop with 3 iterations.
The Return value of the function \(\text{newton}\ (x^3\ \ +\ \ 3x\ \ -\ \ 3,\ \ 3x^2\ \ +\ \ 3,\ \ 1)\) is closest to
- \(083333\)
- \(0.81785\)
- \(0.81773\)
- \(1\)
- \(3\)
Probability, MET2 2023 VCAA 12 MC
The probability mass function for the discrete random variable \(X\) is shown below.
\begin{array} {|c|c|c|c|c|}
\hline X &\ \ \ \ \ \ -1\ \ \ \ \ \ &\ \ \ \ \ \ 0\ \ \ \ \ \ &\ \ \ \ \ \ 1\ \ \ \ \ \ & 2 \\
\hline \text{Pr}(X=x) & k^2 & 3k & k & -k^2-4k+1 \\
\hline \end{array}
The maximum possible value for the mean of \(X\) is:
- \(0\)
- \(\dfrac{1}{3}\)
- \(\dfrac{2}{3}\)
- \(1\)
- \(2\)
Show Worked Solution
| \(E(X)\) | \( =(-1)\times k^2+1\times k+2\times(-k^2-4k+1)\) |
| \(=-3k^2-7k+2\) |
\(\text{Probabilities in table must be}\ \geqslant 0\ \ \Rightarrow \ k \geqslant 0.\)
\(E(X)_{\max}\ \text{occurs when}\ k=0\)
\(E(X)_{\max}\ =2\)
\(\Rightarrow E\)
♦♦ Mean mark 29%.
MARKER’S COMMENT: 26% of students incorrectly chose C.
MARKER’S COMMENT: 26% of students incorrectly chose C.
Functions, MET2 2023 VCAA 9 MC
The function \(f\) is given by
\(f(x) = \begin {cases}
\tan\Bigg(\dfrac{x}{2}\Bigg) &\ \ 4 \leq x \leq 2\pi \\
\sin(ax) &\ \ \ 2\pi\leq x\leq 8
\end{cases}\)
The value of \(a\) for which \(f\) is continuous and smooth at \( x\) = \(2\pi\) is
- \(-2\)
- \(-\dfrac{\pi}{2}\)
- \(-\dfrac{1}{2}\)
- \(\dfrac{1}{2}\)
- \(2\)
Probability, MET2 2023 VCAA 8 MC
A box contains \(n\) green balls and \(m\) red balls. A ball is selected at random, and its colour is noted. The ball is then replaced in the box.
In 8 such selections, where \(n\neq m\), what is the probability that a green ball is selected at least once?
- \(8\Bigg(\dfrac{n}{n+m}\Bigg)\Bigg(\dfrac{m}{n+m}\Bigg)^7\)
- \(1-\Bigg(\dfrac{n}{n+m}\Bigg)^8\)
- \(1-\Bigg(\dfrac{m}{n+m}\Bigg)^8\)
- \(1-\Bigg(\dfrac{n}{n+m}\Bigg)\Bigg(\dfrac{m}{n+m}\Bigg)^7\)
- \(1-8\Bigg(\dfrac{n}{n+m}\Bigg)\Bigg(\dfrac{m}{n+m}\Bigg)^7\)
Calculus, MET2 2023 VCAA 6 MC
Suppose that \(\displaystyle \int_{3}^{10} f(x)\,dx=C\) and \(\displaystyle \int_{7}^{10} f(x)\,dx=D\). The value of \(\displaystyle \int_{7}^{3} f(x)\,dx\) is
- \(C+D\)
- \(C+D-3\)
- \(C-D\)
- \(D-C\)
- \(CD-3\)
Graphs, MET2 2023 VCAA 3 MC
Two function, \(p\) and \(q\), are continuous over their domains, which are \([-2, 3)\) and \((-1, 5]\), respectively.
The domain of the sum function \(p+q\) is
- \([-2, 5]\)
- \([-2, -1)\cup (3, 5]\)
- \([-2, -1)\cup (-1, 3)\cup(3, 5]\)
- \([-1, 3]\)
- \((-1, 3)\)
Algebra, MET2 2023 VCAA 2 MC
For the parabola with equation \(y=ax^2+2bx+c\), where \(a, b, c \in R\), the equation of the axis of symmetry is
- \(x=-\dfrac{b}{a}\)
- \(x=-\dfrac{b}{2a}\)
- \(y=c\)
- \(x=\dfrac{b}{a}\)
- \(x=\dfrac{b}{2a}\)
Graphs, MET1 2022 VCAA 6
The graph of `y=f(x)`, where `f:[0,2 \pi] \rightarrow R, f(x)=2 \sin(2x)-1`, is shown below.
- On the axes above, draw the graph of `y=g(x)`, where `g(x)` is the reflection of `f(x)` in the horizontal axis. (2 marks)
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- Find all values of `k` such that `f(k)=0` and `k \in[0,2 \pi]`. (3 marks)
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- Let `h: D \rightarrow R, h(x)=2 \sin(2x)-1`, where `h(x)` has the same rule as `f(x)` with a different domain.
- The graph of `y=h(x)` is translated `a` units in the positive horizontal direction and `b` units in the positive vertical direction so that it is mapped onto the graph of `y=g(x)`, where `a, b \in(0, \infty)`.
-
- Find the value for `b`. (1 mark)
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- Find the smallest positive value for `a`. (1 mark)
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- Hence, or otherwise, state the domain, `D`, of `h(x)`. (1 mark)
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- Find the value for `b`. (1 mark)
Show Answers Only
a. Graph `y=g(x)`
b. `\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}`
c.i. `b=2`
cii. `=\frac{\pi}{2}`
ciii. `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`
Show Worked Solution
a.
| b. `2 \sin (2 k)-1` | `=0` `0<=k<=2\pi` | |
| `sin (2 k)` | `=1/2` | |
| `2k` | `=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{13 \pi}{6}, \frac{17 \pi}{6}` | |
| `k` | `=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}` |
| c.i ` 2 \sin 2(x-a)-1+b` | `=-(2 \sin 2 x-1)` | |
| `:.\ -1+b` | `=1` | |
| `b` | `=2` |
| c.ii `2 \sin 2(x-a)` | `=-(2 \sin 2 x-1)` | |
| `\sin (2 x-2 a)` | `=-\sin 2 x` | |
| `\therefore 2 a` | `=\pi` | |
| `a` | `=\frac{\pi}{2}` |
♦ Mean mark (c.ii) 50%.
MARKER’S COMMENT: Students confused vertical and horizontal translations. Common error `a=\frac{\pi}{4}.`
MARKER’S COMMENT: Students confused vertical and horizontal translations. Common error `a=\frac{\pi}{4}.`
c.iii The domain for `f(x)` is `[0,2pi]`
`:. \ D` is `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`
♦♦♦ Mean mark (c.iii) 10%.
MARKER’S COMMENT: Common error was translating in the wrong direction. A common incorrect answer was `\left[-\frac{\pi}{2}, \frac{5 \pi}{2}\right].`
MARKER’S COMMENT: Common error was translating in the wrong direction. A common incorrect answer was `\left[-\frac{\pi}{2}, \frac{5 \pi}{2}\right].`
Functions, MET1 2022 VCAA 5b
Find the maximal domain of `f`, where `f(x)=\log _e\left(x^2-2 x-3\right)`. (3 marks)
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Show Worked Solution
`f(x)` exists where `x^2-2 x-3>0`
`:.\ (x – 3)(x + 1) > 0`
From the graph `x < -1` and `x > 3`
`x \in(-\infty,-1) \` ∪ `(3, \infty)`
♦ Mean mark (b) 53%
MARKER’S COMMENT: Students often used incorrect notation. Drawing the graph of the parabola assisted with recognition of correct domain.
MARKER’S COMMENT: Students often used incorrect notation. Drawing the graph of the parabola assisted with recognition of correct domain.
Probability, MET1 2022 VCAA 4
A card is drawn from a deck of red and blue cards. After verifying the colour, the card is replaced in the deck. This is performed four times.
Each card has a probability of `\frac{1}{2}` of being red and a probability of `\frac{1}{2}` of being blue.
The colour of any drawn card is independent of the colour of any other drawn card.
Let `X` be a random variable describing the number of blue cards drawn from the deck, in any order.
- Complete the table below by giving the probability of each outcome. (2 marks)
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- Given that the first card drawn is blue, find the probability that exactly two of the next three cards drawn will be red. (1 mark)
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- The deck is changed so that the probability of a card being red is `\frac{2}{3}` and the probability of a card being blue is `\frac{1}{3}`.
- Given that the first card drawn is blue, find the probability that exactly two of the next three cards drawn will be red. (2 marks)
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Algebra, MET1 2022 VCAA 3
Consider the system of equations
`kx-5y=4+k`
`3x+(k+8) y=-1`
Determine the value of `k` for which the system of equations above has an infinite number of solutions. (3 marks)
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Calculus, MET1 2023 VCAA 9
The shapes of two walking tracks are shown below.
Track 1 is described by the function \(f(x)=a-x(x-2)^2\).
Track 2 is defined by the function \(g(x)=12x-bx^2\).
The unit of length is kilometres.
- Given that \(f(0)=12\) and \(g(1)=9\), verify that \(a=12\) and \(b=-3\). (1 mark)
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- Verify that \(f(x)\) and \(g(x)\) both have a turning point at \(P\).
- Give the co-ordinates of \(P\). (2 marks)
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- A theme park is planned whose boundaries will form the triangle \(\Delta OAB\) where \(O\) is the origin, \(A\) is at \((k, 0)\) and \(B\) is at \((k, g(k))\), as shown below, where \(k \in (0, 4)\).
- Find the maximum possible area of the theme park, in km². (3 marks)
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Calculus, SPEC2 2023 VCAA 9 MC
The position of a particle moving in the Cartesian plane, at time \(t\), is given by the parametric equations
\(x(t)=\dfrac{6 t}{t+1}\) and \(y(t)=\dfrac{-8}{t^2+4}\), where \(t \geq 0\).
What is the slope of the tangent to the path of the particle when \(t=2\) ?
- \(-\dfrac{1}{3}\)
- \(-\dfrac{1}{4}\)
- \(\dfrac{1}{3}\)
- \(\dfrac{3}{4}\)
- \(\dfrac{4}{3}\)
Calculus, MET1 2023 VCAA 7
Consider \(f:(-\infty, 1]\rightarrow R, f(x)=x^2-2x\). Part of the graph of \(y=f(x)\) is shown below.
- State the range of \(f\). (1 mark)
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- Sketch the graph of the inverse function \(y=f^{-1}(x)\) on the axes above. Label any endpoints and axial intercepts with their coordinates. (2 marks)
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- Determine the equation of the domain for the inverse function \(f^{-1}\). (2 marks)
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- Calculate the area of the regions enclosed by the curves of \(f,\ f^{-1}\) and \(y=-x\). (2 marks)
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Show Answers Only
a. \([-1, \infty)\)
b.
c. \(f^{-1}(x)=1-\sqrt{x+1}\)
\(\text{Domain}\ [-1, \infty)\)
d. \(A=\dfrac{1}{3}\)
Show Worked Solution
a. \([-1, \infty)\)
b.
c. \(\text{When }f(x)\ \text{is written in turning point form}\)
\(y=(x-1)^2-1\)
\(\text{Inverse function: swap}\ x \leftrightarrow y\)
| \(x\) | \(=(y-1)^2-1\) |
| \(x+1\) | \(=(y-1)^2\) |
| \(-\sqrt{x+1}\) | \(=y-1\) |
| \(f^{-1}(x)\) | \(=1-\sqrt{x+1}\) |
\(\text{Domain}\ [-1, \infty)\)
♦ Mean mark (c) 48%.
MARKER’S COMMENT: Common error → writing the function as \(f^{-1}(x)=1+\sqrt{x+1}\).
MARKER’S COMMENT: Common error → writing the function as \(f^{-1}(x)=1+\sqrt{x+1}\).
d. \(\text{One strategy of many possibilities:}\)
| \(A\) | \(=2\displaystyle \int_{0}^{1} \big(-x-(x^2-2x)\big)\,dx\) | |
| \(=2\displaystyle \int_{0}^{1} \big(x-x^2\big)\,dx\) | ||
| \(=2\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1\) | ||
| \(=2\bigg(\dfrac{1}{2}-\dfrac{1}{3}-(0)\bigg)\) | ||
| \(=\dfrac{1}{3}\) |
♦♦♦ Mean mark (d) 24%.
MARKER’S COMMENT: Using the symmetry properties of the graph and its inverse helped answer this question efficiently.
MARKER’S COMMENT: Using the symmetry properties of the graph and its inverse helped answer this question efficiently.
Statistics, MET1 2023 VCAA 6
Let \(\hat{P}\) be the random variable that represents the sample proportion of households in a given suburb that have solar panels installed.
From a sample of randomly selected households in a given suburb, an approximate 95% confidence interval for the proportion \(p\) of households having solar panels installed was determined to be (0.04, 0.16).
- Find the value of \(\hat{p}\) that was used to obtain this approximate 95% confidence interval. (1 mark)
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Use \(z=2\) to approximate the 95% confidence interval.
- Find the size of the sample from which this 95% confidence interval was obtained. (2 marks)
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- A larger sample of households is selected, with a sample size four times the original sample.
The sample proportion of households having solar panels installed is found to be the same. - By what factor will the increased sample size affect the width of the confidence interval? (1 mark)
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Probability, MET1 2023 VCAA 8
Suppose that the queuing time, \(T\) (in minutes), at a customer service desk has a probability density function given by
\(f(t) = \begin {cases}
kt(16-t^2) &\ \ 0 \leq t \leq 4 \\
\\
0 &\ \ \text{elsewhere}
\end{cases}\)
for some \(K \in R\).
- Show that \(k=\dfrac{1}{64}\). (1 mark)
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- Find \(\text{E}(T)\). (2 marks)
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- What is the probability that a person has to queue for more than two minutes, given that they have already queued for one minute? (3 marks)
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Calculus, MET1 2022 VCAA 2b
Evaluate `\int_0^1(f(x)(2 f(x)-3))dx`, where `\int_0^1[f(x)]^2 dx=\frac{1}{5}` and `\int_0^1 f(x) dx=\frac{1}{3}`. (3 marks)
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Calculus, MET1 2022 VCAA 2a
Let `g:\left(\frac{3}{2}, \infty\right) \rightarrow R, g(x)=\frac{3}{2 x-3}`
Find the rule for an antiderivative of `g(x)`. (1 mark)
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Graphs, MET1 2023 VCAA 3
- Sketch the graph of \(f(x)=2-\dfrac{3}{x-1}\) on the axes below, labelling all asymptotes with their equation and axial intercepts with their coordinates. (3 marks)
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- Find the values of \(x\) for which \(f(x)\leq1\). (1 mark)
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Show Answers Only
a.
b. \(1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]\)
Show Worked Solution
a. \(\text{Vertical asymptote when}\ \ x=1\)
\(y\text{-int:}\ y=2-(-3)\ \ \Rightarrow\ \ y=5\)
\(x\text{-int:}\ 2-\dfrac{3}{x-1}=0\ \ \Rightarrow\ \ x=\dfrac{5}{2} \)
\(\text{As}\ \ x \rightarrow \infty, \ \ y \rightarrow 2^{-}; \ \ x \rightarrow -\infty, \ \ y \rightarrow 2^{+} \)
b. \(\text{From the graph:}\)
\(f(x)=1\ \text{when }x=4\)
\(x>1\ \text{to the right of the vertical asymptote}\)
\(\therefore\ f(x)\leq1\ \text{when}\ \ 1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]\)
♦♦ Mean mark (b) 38%.
MARKER’S COMMENT: Many students did not use their graph from part (a).
MARKER’S COMMENT: Many students did not use their graph from part (a).
Vectors, SPEC1 2023 VCAA 10
The position vector of a particle at time \(t\) seconds is given by
\(\underset{\sim}{\text{r}}(t)=\big{(}5-6 \ \sin ^2(t) \big{)} \underset{\sim}{\text{i}}+(1+6 \ \sin (t) \cos (t)) \underset{\sim}{\text{j}}\), where \(t \geq 0\).
- Write \(5-6\, \sin ^2(t)\) in the form \(\alpha+\beta\, \cos (2 t)\), where \(\alpha, \beta \in Z^{+}\). (1 mark)
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- Show that the Cartesian equation of the path of the particle is \((x-2)^2+(y-1)^2=9.\) (2 marks)
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- The particle is at point \(A\) when \(t=0\) and at point \(B\) when \(t=a\), where \(a\) is a positive real constant.
- If the distance travelled along the curve from \(A\) to \(B\) is \(\dfrac{3 \pi}{4}\), find \(a\). (1 mark)
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- Find all values of \(t\) for which the position vector of the particle, \(\underset{\sim}{\text{r}}(t)\), is perpendicular to its velocity vector, \(\underset{\sim}{\dot{\text{r}}}(t)\). (2 marks)
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Vectors, SPEC1 2023 VCAA 9
A plane contains the points \( A(1,3,-2), B(-1,-2,4)\) and \( C(a,-1,5)\), where \(a\) is a real constant. The plane has a \(y\)-axis intercept of 2 at the point \(D\).
- Write down the coordinates of point \(D\). (1 mark)
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- Show that \(\overrightarrow{A B}\) and \(\overrightarrow{A D}\) are \(-2 \underset{\sim}{\text{i}}-5 \underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\) and \(-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2 \underset{\sim}{\text{k}}\), respectively. (1 mark)
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- Hence find the equation of the plane in Cartesian form. (2 marks)
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- Find \(a\). (1 mark)
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- \(\overline{A B}\) and \(\overline{A D}\) are adjacent sides of a parallelogram. Find the area of this parallelogram. (1 mark)
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Calculus, SPEC1 2023 VCAA 7
The curve defined by the parametric equations
\(x=\dfrac{t^2}{4}-1, \ y=\sqrt{3} t\), where \(0 \leq t \leq 2 \text {, }\)
is rotated about the \(x\)-axis to form an open hollow surface of revolution.
Find the surface area of the surface of revolution.
Give your answer in the form \(\pi\left(\dfrac{a \sqrt{b}}{c}-d\right)\), where \(a, b, c\) and \(d \in Z^{+}\).
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Networks, GEN2 2023 VCAA 14
One of the landmarks in state \(A\) requires a renovation project.
This project involves 12 activities, \(A\) to \(L\). The directed network below shows these activities and their completion times, in days.
The table below shows the 12 activities that need to be completed for the renovation project.
It also shows the earliest start time (EST), the duration, and the immediate predecessors for the activities.
The immediate predecessor(s) for activity \(I\) and the EST for activity \(J\) are missing.
\begin{array} {|c|c|c|}
\hline
\quad \textbf{Activity} \quad & \quad\quad\textbf{EST} \quad\quad& \quad\textbf{Duration}\quad & \textbf{Immediate} \\
& & & \textbf{predecessor(s)} \\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & 0 & 6 & - \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 0 & 4 & - \\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 6 & 7 & A \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & 4 & 5 & B \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & 4 & 10 & B \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & 13 & 4 & C \\
\hline
\rule{0pt}{2.5ex} G \rule[-1ex]{0pt}{0pt} & 9 & 3 & D \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & 9 & 7 & D \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 13 & 6 & - \\
\hline
\rule{0pt}{2.5ex} J \rule[-1ex]{0pt}{0pt} & - & 6 & E, H \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & 19 & 4 & F, I \\
\hline
\rule{0pt}{2.5ex} L \rule[-1ex]{0pt}{0pt} & 23 & 1 & J, K \\
\hline
\end{array}
- Write down the immediate predecessor(s) for activity \(I\). (1 mark)
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- What is the earliest start time, in days, for activity \(J\) ? (1 mark)
--- 1 WORK AREA LINES (style=lined) ---
- How many activities have a float time of zero? (1 mark)
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The managers of the project are able to reduce the time, in days, of six activities.
These reductions will result in an increase in the cost of completing the activity.
The maximum decrease in time of any activity is two days.
\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Activity} \rule[-1ex]{0pt}{0pt} & \quad \quad A \quad \quad & \quad \quad B \quad \quad& \quad \quad F \quad \quad & \quad \quad H \quad \quad & \quad \quad I \quad \quad & \quad \quad K \quad \quad \\
\hline
\rule{0pt}{2.5ex} \textbf{Daily cost (\$)} \rule[-1ex]{0pt}{0pt} & 1500 & 2000 & 2500 & 1000 & 1500 & 3000 \\
\hline
\end{array}
- If activities \(A\) and \(B\) have their completion time reduced by two days each, the overall completion time of the project will be reduced.
- What will be the maximum reduction time, in days? (1 mark)
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- The managers of the project have a maximum budget of $15 000 to reduce the time for several activities to produce the maximum reduction in the project's overall completion time.
- Complete the table below, showing the reductions in individual activity completion times that would achieve the earliest completion time within the $ 15 000 budget. (1 mark)
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\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}
Show Worked Solution
a. \(\text{Immediate predecessors of}\ I:\ C, G\)
\(\text{Dummy activity before activity}\ C\ \text{does not effect this.}\)
b. \(\text{Scan network:}\)
\(\text{EST}(J) = 4+5+7 = 16\ \text{days}\)
c. \(\text{Critical Path:}\ A\ C\ I\ K\ L\)
\(\text{Activities on the critical path have a float time of zero.}\)
\(\Rightarrow \ \text{5 activities have a float time of zero.}\)
d. \(\text{If activities}\ A\ \text{and}\ B\ \text{are reduced by 2 days,}\)
\(\text{the critical path remains:}\ A\ C\ I\ K\ L\ \text{(22 days)}\)
\(\text{Maximum reduction time = 2 days}\)
♦♦ Mean mark (c) 35%.
♦♦ Mean mark (d) 33%.
♦♦ Mean mark (d) 33%.
e.
\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & 0\\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 2\\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & 1\\
\hline
\end{array}
♦♦♦ Mean mark (e) 9%.
Networks, GEN2 2023 VCAA 13
The state \(A\) has nine landmarks, \(G, H, I, J, K, L, M, N\) and \(O\).
The edges on the graph represent the roads between the landmarks.
The numbers on each edge represent the length, in kilometres, along each road.
Three friends, Eden, Reynold and Shyla, meet at landmark \(G\).
- Eden would like to visit landmark \(M\).
- What is the minimum distance Eden could travel from \(G\) to \(M\) ? (1 mark)
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- Reynold would like to visit all the landmarks and return to \(G\).
- Write down a route that Reynold could follow to minimise the total distance travelled. (1 mark)
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- Shyla would like to travel along all the roads.
- To complete this journey in the minimum distance, she will travel along two roads twice.
- Shyla will leave from landmark \(G\) but end at a different landmark.
- Complete the following by filling in the boxes provided.
- The two roads that will be travelled along twice are the roads between: (1 mark)
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Matrices, GEN2 2023 VCAA 11
The circus requires 180 workers to put on each show. From one show to the next, workers can either continue working \((W)\) or they can leave the circus \((L)\). Once workers leave the circus, they do not return. It is known that 95% of the workers continue working at the circus. This situation can be modelled by the matrix recurrence relation \(S_0=\begin{bmatrix}180\\ 0\end{bmatrix}, \quad \quad S_{n+1}=T S_n+B\) --- 0 WORK AREA LINES (style=lined) --- \({\displaystyle} --- 0 WORK AREA LINES (style=lined) --- \({\displaystyle}
\begin{aligned}
& \quad \quad\quad \ \ \ \textit{this show}\\
& \quad \quad \quad \ \ \ W \quad \quad L \\
& T=\begin{bmatrix}
\ \rule[-3ex]{1cm}{0.15mm} & \ \rule[-3ex]{1cm}{0.15mm} \ \\
\ \rule[-3ex]{1cm}{0.15mm} & \ \rule[-3ex]{1cm}{0.15mm} \ \\
\rule[1ex]{0pt}{0pt}
\end{bmatrix} \begin{array}{ll}
&\rule[0ex]{0pt}{0pt}\\
\rule[-3ex]{0pt}{0pt}W\\
\rule[-3ex]{0pt}{0pt}L
\end{array} \ \textit{ next show}
&
\end{aligned}\)
\begin{aligned}
& B=\begin{bmatrix}
\ \rule[-3ex]{1cm}{0.15mm}\ \\
\ \rule[-3ex]{1cm}{0.15mm} \ \\
\rule[1ex]{0pt}{0pt}
\end{bmatrix}
&
\end{aligned}\)
Matrices, GEN2 2023 VCAA 10
Within the circus, there are different types of employees: directors \((D)\), managers \((M)\), performers \((P)\) and sales staff \((S).\) Customers \((C)\) attend the circus. Communication between the five groups depends on whether they are customers or employees, and on what type of employee they are. Matrix \(G\) below shows the communication links between the five groups. \begin{aligned} In this matrix: --- 2 WORK AREA LINES (style=lined) --- \begin{aligned} --- 0 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) ---
&\quad \quad \quad\quad \quad \quad\quad \quad \quad \ \ \textit{receiver}\\
&\quad \quad\quad \quad \quad\quad \quad \quad D \ \ M \ \ P \ \ \ S \ \ \ C \\
& G=\textit{sender} \quad \begin{array}{ccccc}
D\\
M\\
P\\
S\\
C
\end{array}
\begin {bmatrix}
0 & 1 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0
\end{bmatrix}\\
&
\end{aligned}
&\quad \quad \quad\quad \quad \quad\quad \quad \quad \ \ \textit{receiver}\\
&\quad \quad\quad \quad \quad\quad \quad \quad D \quad M \quad P \quad \ S \quad \ C \\
& H=\textit{sender} \quad \begin{array}{ccccc}
D\\
M\\
P\\
S\\
C
\end{array}
\begin {bmatrix} {\displaystyle}
1 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
0 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
1 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
1 & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} & \rule{0.5cm}{0.15mm} \\
0 & 1 & 0 & 0 & 1
\end{bmatrix}\\
&
\end{aligned}
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