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v1 Algebra, STD2 A1 SM-Bank 7

If  \(S = V_0(1 - r)^n\), find \(S\) given  \(V_0 = $57\ 000\), \(r = 0.12\) and \(n=5\). (give your answer to the nearest cent).  (2 marks)

Show Answers Only

\($30\ 080.72\ \text{(to nearest cent)}\)

Show Worked Solution
\(S\) \(=V_0(1 – r)^n\)
  \(=57\ 000 (1-0.12)^5\)
  \(=57\ 000 (0.88)^5\)
  \(=$30\ 080.719\dots\)
  \(=$30\ 080.72\ \text{(to nearest cent)}\)

EXAMCOPY Functions, MET2 2022 VCAA 4

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

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

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

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

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

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

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

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

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

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

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

a.   `R` is the range.

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

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

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

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

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

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

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

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

`\frac{1}{k}(4)<1`   [Using CAS]

`:.\  k > 4`


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

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

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


♦♦♦♦ Mean mark (e.ii) 10%.

CHEMISTRY, M8 2013 VCE 8* MC

A forensic chemist tests mud from a crime scene to determine whether the mud contains zinc. Which one of the following analytical techniques would be best suited to this task?

  1. infrared spectroscopy
  2. mass spectroscopy
  3. atomic absorption spectroscopy
  4. nuclear magnetic resonance spectroscopy
Show Answers Only

\(C\)

Show Worked Solution
  • AAS is the only analytical technique that is specific to identifying metals within a solution.

\(\Rightarrow C\)

CHEMISTRY, M7 2018 VCE 1a

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

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

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

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

Show Answers Only

i.    
         

ii.    \(\ce{C4H6}\)

  • Structural or semi-structural formulas are not appropriate.
     

iii. •   Longest carbon chain is 6, hence hexane

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

Show Worked Solution

i.    
         

ii.    \(\ce{C4H6}\)

  • Structural or semi-structural formulas are not appropriate.
     

iii.    Longest carbon chain is 6, hence hexane

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

CHEMISTRY, M4 2012 VCE 5 MC

Nitrogen dioxide decomposes as follows.

\(\ce{2NO2(g) \rightarrow N2(g) + 2O2(g)}\ \quad \quad \Delta H = -66 \text{ kJ mol}^{-1}\)

The enthalpy change for the reaction represented by the equation  \(\ce{\frac{1}{2}N2(g) + O2(g) \rightarrow NO2(g)}\)  is

  1. \(-66 \text{ kJ mol} ^{-1}\)
  2. \(-33 \text{ kJ mol} ^{-1}\)
  3. \(+33 \text{ kJ mol} ^{-1}\)
  4. \(+66 \text{ kJ mol} ^{-1}\)
Show Answers Only

\(C\)

Show Worked Solution

\(\ce{2NO2(g) \rightarrow N2(g) + 2O2(g),}\ \quad \Delta H = -66 \text{ kJ mol}^{-1}\)

\(\ce{N2(g) + 2O2(g) \rightarrow 2NO2(g),}\ \quad \Delta H = +66 \text{ kJ mol}^{-1}\)

\(\ce{\frac{1}{2}N2(g) + O2(g) \rightarrow NO2(g),}\ \quad \Delta H = +33 \text{ kJ mol}^{-1}\)

\(\Rightarrow C\)

CHEMISTRY, M4 2013 VCE 16*

\(\ce{C(s) + O2(g)\rightarrow CO2(g)}\) \(\quad \quad \Delta H = -393.5 \text{ kJ mol}^{-1}\)
\(\ce{2H(g) + O2(g)\rightarrow 2H2O(l)}\) \(\quad \quad \Delta H = -571.6 \text{ kJ mol}^{-1}\)

Given the information above, what is the enthalpy change for the following reaction?   (2 marks)

\(\ce{C(s) + 2H2O(l)\rightarrow CO2(g) + 2H2(g)}\)

Show Answers Only

\(\Delta H = 178.1\ \text{kJ mol}^{-1}\)

Show Worked Solution

1st reaction is unchanged:

\(\ce{C(s) + O2(g)\rightarrow CO2(g)}\quad \Delta H = -393.5 \text{ kJ mol}^{-1}\)
 

Reverse 2nd reaction:

\(\ce{2H2O(l) \rightarrow 2H(g) + O2(g) }\quad \Delta H = +571.6 \text{ kJ mol}^{-1}\)
 

Combine both reactions:

\(\ce{C(s) + 2H2O(l)\rightarrow CO2(g) + 2H2(g)}\)

\(\Delta H = -393.5+571.6 = 178.1\ \text{kJ mol}^{-1}\)

CHEMISTRY, M4 2016 VCE 17*

The combustion of hexane takes place according to the equation

\(\ce{C6H14(g) + \dfrac{19}{2}O2(g)\rightarrow 6CO2(g) + 7H2O(g)}\) \(\quad \quad \Delta H = -4158\ \text{kJ mol}^{-1}\)

Consider the following reaction.

\(\ce{ 12CO2(g) + 14H2O(g)\rightarrow 2C6H14(g) + 19O2(g)}\)

  1. Calculate the value of \(\Delta H\), in kJ mol\(^{-1}\), for this reaction   (2 marks)

  2. Is the reaction exothermic or endothermic?   (1 mark)

Show Answers Only

a.    \(\Delta H = +8316\ \text{kJ mol}^{-1}\)

b.    Endothermic

Show Worked Solution

a.    \(\ce{C6H14(g) + \dfrac{19}{2}O2(g)\rightarrow 6CO2(g) + 7H2O(g)}\) \(\quad \Delta H = -4158\ \text{kJ mol}^{-1}\)

  • Reverse equation:
  •   \(\ce{6CO2(g) + 7H2O(g)\rightarrow C6H14(g) + \dfrac{19}{2}O2(g)}\) \(\quad \Delta H = +4158\ \text{kJ mol}^{-1}\)
  • Double equation:
  •   \(\ce{12CO2(g) + 14H2O(g)\rightarrow 2C6H14(g) + 19O2(g)}\) \(\quad \Delta H = 2 \times +4158\ \text{kJ mol}^{-1}\)
  • \(\Delta H = +8316\ \text{kJ mol}^{-1}\)

b.    Endothermic

CHEMISTRY, M4 2015 VCE 17 MC

The oxidation of sulfur dioxide is an exothermic reaction. The reaction is catalysed by vanadium\(\text{(V)}\) oxide.

\(\ce{2SO2(g) + O2(g)\rightleftharpoons 2SO3(g)}\)

Which one of the following energy profile diagrams correctly represents both the catalysed and the uncatalysed reaction?
 


 

Show Answers Only

\(B\)

Show Worked Solution

→ Exothermic reaction releases energy (eliminate C and D).

→ Catalysts only effect the enthalpy during the chemical reaction by reducing the activation energy required.

\(\Rightarrow B\)

CHEMISTRY, M4 2018 VCE 14*

An equation for the complete combustion of methanol is

\(\ce{2CH3OH(l) + 3O2(g)\rightarrow 2CO2(g) + 4H2O(g)}\ \ \ \ \ \ \ \Delta H=-726\ \text{kJ mol}^{-1}\)

  1. State whether this reaction exothermic or endothermic.   (1 mark)

  2. Calculate the total enthalpy change of the equation, in kilojoules.   (1 mark)

Show Answers Only

a.    Methanol combustion is an exothermic reaction.

\(\Delta H=-726\ \text{kJ mol}^{-1}\)  also indicates it is exothermic.
 

b.    \(\Delta H\ \ce{(CH3OH) = 726\ \text{kJ mol}^{-1}}\)

  • 2 moles of \(\ce{CH3OH}\) are involved.
  • \(\text{Total}\ \Delta H = 2 \times -726 = -1452\ \text{kJ}\)
Show Worked Solution

a.    Methanol combustion is an exothermic reaction.

\(\Delta H=-726\ \text{kJ mol}^{-1}\)  also indicates it is exothermic.
 

b.    \(\Delta H\ \ce{(CH3OH) = 726\ \text{kJ mol}^{-1}}\)

  • 2 moles of \(\ce{CH3OH}\) are involved.
  • \(\text{Total}\ \Delta H = 2 \times -726 = -1452\ \text{kJ}\)

PHYSICS, M5 2019 VCE 10

A projectile is launched from the ground at an angle of 39° and at a speed of 25 m s\(^{-1}\), as shown in the diagram. The maximum height that the projectile reaches above the ground is labelled \(h\).
 

  1. Ignoring air resistance, show that the projectile's time of flight from the launch to the highest point is equal to 1.6 seconds. Give your answer to two significant figures. Show your working and indicate your reasoning.  (2 marks)

  2. Calculate the range, \(R\), of the projectile. Show your working.  (2 marks)

Show Answers Only

a.    \(t=1.6\ \text{s}\)

b.    \(62\ \text{m}\)

Show Worked Solution

a.    
       

\(v_v=25\,\sin39^{\circ}=15.733\ \text{ms}^{-1}\)

\(\text{At max height:}\ v_v=0\)

\(v\) \(=u+at_1\)  
\(0\) \(=15.733 -9.8t_1\)  
\(9.8t_1\) \(=15.733\)  
\(t_1\) \(=1.6\ \text{s  (2 sig.fig)}\)  

 

b.    \(\text{Since path is symmetrical:}\)

\(\text{Time of flight}\ (t_2) =1.6 \times 2=3.2\ \text{s}\).

\(v_h=25\,\cos39^{\circ}=19.43\ \text{ms}^{-1}\ \ \text{(see part (a) diagram)}\) 

\(\therefore R=v_ht_2=19.43 \times 3.2= 62\ \text{m}\)

PHYSICS, M6 2019 VCE 7*

Students in a Physics practical class investigate the piece of electrical equipment shown in the diagram. It consists of a single rectangular loop of wire that can be rotated within a uniform magnetic field. The loop has dimensions 0.50 m × 0.25 m and is connected to the output terminals with slip rings. The loop is in a uniform magnetic field of strength 0.40 T.
 

  1. What name best describes the piece of electrical equipment shown in the diagram?   (1 mark)

  2. What is the magnitude of the flux through the loop when it is in the position shown in the diagram? Explain your answer.   (2 marks)

The students connect the output terminals of the piece of electrical equipment to an oscilloscope. One student rotates the loop at a constant rate of 20 revolutions per second.

  1. Calculate the period of the rotation of the loop.   (1 mark)

  2. Calculate the maximum flux through the loop. Show your working.   (1 mark)

  3. The loop starts in the position shown in the diagram.
  4. What is the average voltage measured across the output terminals for the first quarter turn? Show your working.   (2 marks)

Show Answers Only

a.    Alternator.

b.   \(0\ \text{Wb}\)

  • This is because the plane of the area of the coil is parallel to the direction of the magnetic field, not perpendicular.

c.    \(0.05\ \text{s}\)

d.    \(0.05\ \text{Wb}\)

e.    \(4\ \text{V}\)

Show Worked Solution

a.    The device is an alternator.

  • This is due to the input being mechanical motion and the output being AC current, whereas an AC motor is the opposite.
♦ Mean mark (a) 44%.
COMMENT: Many students incorrectly answered AC motor.

b.   \(0\ \text{Wb}\)

  • This is because the plane of the area of the coil is parallel to the direction of the magnetic field, not perpendicular.

c.    \(T=\dfrac{1}{f}=\dfrac{1}{20}=0.05\ \text{s}\)
 

d.    \(\Phi=BA=0.4 \times (0.5 \times 0.25)=0.05\ \text{Wb}\)
 

e.     \(\varepsilon\) \(=\dfrac{\Delta \Phi}{\Delta t_{\text{1/4 rotation}}}\)
    \(=\dfrac{0.05}{0.0125}\)
    \(=4\ \text{V}\)
♦ Mean mark (e) 51%.

PHYSICS, M6 2019 VCE 6

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

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

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

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

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

Show Answers Only

a.    \(12\ \text{W}\)

b.   Potential cause of dimmer lights: 

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

c.    Changes using the same equipment:

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

Show Worked Solution

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

\(P=IV=1 \times 12 = 12\ \text{W}\)
 

b.   Potential cause of dimmer lights: 

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

c.    Changes using the same equipment:

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

PHYSICS, M6 2019 VCE 3

The diagram below shows a schematic diagram of a DC motor. The motor has a coil, \(JKLM\), consisting of 100 turns. The permanent magnets provide a uniform magnetic field of 0.45 T.

The commutator connectors, \(X\) and \(Y\), provide a constant DC current, \(I\), to the coil. The length of the side \(JK\) is 5.0 cm.

The current \(I\) flows in the direction shown in the diagram.
 

  1. Which terminal of the commutator is connected to the positive terminal of the current supply?   (1 mark)

  2. Draw an arrow on the diagram to indicate the direction of the magnetic force acting on the side \(JK\).   (1 mark)

  3. Explain the role of the commutator in the operation of the DC motor.   (2 marks)

  4. A current of 6.0 A flows through the 100 turns of the coil \(JKLM\).
  5. The side \(JK\) is 5.0 cm in length.
  6. Calculate the size of the magnetic force on the side \(JK\) in the orientation shown in Figure 3. Show your working.   (2 marks)

Show Answers Only

a.    Current runs from the positive terminal to the negative terminal.

\(X\) is connected to the positive terminal of the current supply.
 

b.    Using the right-hand rule: force on \(JK\) is down the page.
 

c.    Role of the split-ring commutator:

  • Changes the direction of the current through the arms of the coil every 180\(^{\circ}\).
  • This ensures that the torque through the motor will always be in the same direction so the motor will rotate in the same direction.

d.    \(F=nlIB=100 \times 0.05 \times 6 \times 0.45 = 13.5\ \text{N}\)

Show Worked Solution

a.    Current runs from the positive terminal to the negative terminal.

\(X\) is connected to the positive terminal of the current supply.
 

b.    Using the right-hand rule: force on \(JK\) is down the page.
 

c.    Role of the split-ring commutator:

  • Changes the direction of the current through the arms of the coil every 180\(^{\circ}\).
  • This ensures that the torque through the motor will always be in the same direction so the motor will rotate in the same direction.

d.    \(F=nlIB=100 \times 0.05 \times 6 \times 0.45 = 13.5\ \text{N}\)

PHYSICS, M7 2019 VCE 16 MC

Students are conducting a photoelectric effect experiment. They shine light of known frequency onto a metal and measure the maximum kinetic energy of the emitted photoelectron.

The students increase the intensity of the incident light.

The effect of this increase would most likely be

  1. lower maximum kinetic energy of the emitted photoelectrons.
  2. higher maximum kinetic energy of the emitted photoelectrons.
  3. fewer emitted photoelectrons but of higher maximum kinetic energy.
  4. more emitted photoelectrons but of the same maximum kinetic energy.
Show Answers Only

\(D\)

Show Worked Solution
  • The energy of a photon (which is transferred to the photoelectron) is independent of the intensity of the light.
  • Increasing the intensity of light will not change the maximum kinetic energy of the photoelectrons.
  • However, a higher intensity of light corresponds to more photons and thus more emitted photoelectrons.

\(\Rightarrow D\)

PHYSICS, M8 2019 VCE 14* MC

Electrons are accelerated in an electron gun to a speed of 1.0 × 10\(^7\) m s\(^{-1}\).

The best estimate of the de Broglie wavelength of these electrons is

  1. 4.5 × 10\(^{-6}\) m
  2. 7.3 × 10\(^{-8}\) m
  3. 7.3 × 10\(^{-11}\) m
  4. 4.5 × 10\(^{-12}\) m
Show Answers Only

\(C\)

Show Worked Solution

\(\lambda=\dfrac{h}{mv}=\dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 1 \times 10^7}=7.3 \times 10^{-11}\ \text{m}\)

\(\Rightarrow C\)

PHYSICS, M6 2019 VCE 5-6* MC

A 40 V AC generator and an ideal transformer are used to supply power. The diagram below shows the generator and the transformer supplying 240 V to a resistor with a resistance of 1200 \( \Omega \).
 

Question 5

Which of the following correctly identifies the parts labelled \(\text{X}\) and \(\text{Y}\), and the function of the transformer?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|l|l|l|}
\hline
\rule{0pt}{2.5ex}\quad \ \ \ \text{Part X}\quad \rule[-1ex]{0pt}{0pt}&\ \ \  \quad \text{Part Y} \quad& \text{Function of transformer} \\
\hline
\rule{0pt}{2.5ex}\text{primary coil}\rule[-1ex]{0pt}{0pt}&\text{secondary coil} & \text{step-down}\\
\hline
\rule{0pt}{2.5ex}\text{primary coil}\rule[-1ex]{0pt}{0pt}& \text{secondary coil}&\text{step-up}\\
\hline
\rule{0pt}{2.5ex}\text{secondary coil}\rule[-1ex]{0pt}{0pt}& \text{primary coil} &\text{step-down}\\
\hline
\rule{0pt}{2.5ex}\text{secondary coil}\rule[-1ex]{0pt}{0pt}& \text{primary coil} &\text{step-up}\\
\hline
\end{array}
\end{align*}

 
Question 6

Which one of the following is closest to the current in the primary circuit?

  1. \(0.04\ \text{A}\)
  2. \(0.20\ \text{A}\)
  3. \(1.20\ \text{A}\)
  4. \(1.50\ \text{A}\)
Show Answers Only

\(\text{Question 5:}\ B\)

\(\text{Question 6:}\ C\)

Show Worked Solution

Question 5

  • The primary coil is connected to the power supply (i.e. Part \(\text{X}\)).
  • For a step-up transformer, there are more coils in the secondary coil then there are in the primary coil.

\(\Rightarrow B\)
 

Question 6

\(I_s=\dfrac{V_s}{r_s}=\dfrac{240}{1200}=0.2\ \text{A}\)

\(\dfrac{I_s}{I_p}\) \(=\dfrac{N_p}{N_s}\)  
\(I_p\) \(=\dfrac{N_s \times I_s}{N_p}=\dfrac{6000 \times 0.2}{1000}=1.2\ \text{A}\)  

 

\(\Rightarrow C\)

Mean mark Q6 57%.

Calculus, MET1 2023 VCAA SM-Bank 6

Newton's method is used to estimate the \(x\)-intercept of the function  \(f(x)=\dfrac{1}{3} x^3+2 x+4\).

  1. Verify that \(f(-1)>0\) and \(f(-2)<0\).  (1 mark)

  2. Using an initial estimate of \(x_0=-1\), find the value of \(x_1\).  (2 marks)

Show Answers Only

a.    \(\text{See worked solutions}\)

b.    \(-\dfrac{14}{9}\)

Show Worked Solution
a.     \(f(x)\) \(=\dfrac{1}{3}x^3+2x+4\)
  \(f(-1)\) \(=\dfrac{1}{3}.(-1)^3+2.(-1)+4=\dfrac{5}{3}>0\)
  \(f(-2)\) \(=\dfrac{1}{3}.(-2)^3+2.(-2)+4=-\dfrac{8}{3}<0\)

 

b.     \(x_1\) \(=x_0-\dfrac{f(x_0)}{f^{\prime}(x_0)},\quad\ \text{where }x_0=-1\)
   

\(=-1-\dfrac{-\dfrac{1}{3}+2.(-1)+4}{(-1)^2+2}\)
    \(=-\dfrac{14}{9}\)

Functions, MET1 EQ-Bank 2

Consider the functions \(f\) and \(g\), where

\begin{aligned}
& f: R \rightarrow R, f(x)=x^2-9 \\
& g:[0, \infty) \rightarrow R, g(x)=\sqrt{x}
\end{aligned}

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

  2. Determine the rule for the equation and state the domain of the function \(f \circ g\).  (2 marks)

  3. Let \(h\) be the function \(h: D \rightarrow R, h(x)=x^2-9\).
  4. Determine the maximal domain, \(D\), such that \(g \circ h\) exists.  (1 marks)

Show Answers Only

a.    \([-9, \infty)\)

b.    \(f\circ g(x)=x-9, \text{Domain}\ [0, \infty)\)

c.    \((-\infty, -3)\cap (3, \infty)\)

Show Worked Solution

a.    \(\text{Range}\ \rightarrow\  [-9, \infty)\)

b.     \(f\circ g(x)\) \(=(g(x))^2-9\)
    \(=(\sqrt{x})^2-9\)
    \(=x-9\)

\(g(x)=\sqrt{x} \ \rightarrow x\ \text{must be }\geq 0\)

\(\therefore\ \text{Domain}\ f\circ g(x) \text{ is }[0, \infty)\)

c.     \(g\circ h(x)\) \(=\sqrt{h(x)}\)
    \(=\sqrt{x^2-9}\)

\(\text{For }g\circ h(x)\ \text{to exist}\ h(x)\geq 0\)

\(x\text{-intercepts for }h(x)\ \text{are } x=-3, 3\)

\(\text{and }h(x)\ \text{is positive for } x\leq -3\ \text{and }x\geq 3\)

\(\therefore\ \text{Maximal domain} = (-\infty, -3)\cap (3, \infty)\)

Graphs, MET1 EQ-Bank 1

Let  \(\displaystyle f:[-3,-2) \cup(-2, \infty) \rightarrow R, f(x)=1+\frac{1}{x+2}\).

  1. On the axes below, sketch the graph of \(f\). Label any asymptotes with their equations, and endpoints and axial intercepts with their coordinates.   (3 marks)

  1. Find the values of \(x\) for which \(f(x) \leq 2\).   (2 marks)

Show Answers Only

a.
   

b.   \(x\in [-3, -2)\cap (-1, \infty)\)

Show Worked Solution

a.
     

b.    \(\text{From graph }f(x)\leq -2\ \text{ for}\ -3\leq x <2\ \text{ and when }\ x\geq -1\) 

\(\rightarrow x\in [-3, -2)\cap (-1, \infty)\)

Calculus, MET2 EQ-Bank 2

Jac and Jill have built a ramp for their toy car. They will release the car at the top of the ramp and the car will jump off the end of the ramp.

The cross-section of the ramp is modelled by the function \(f\), where

\(f(x)= \begin{cases}\displaystyle \ 40 & 0 \leq x<5 \\ \dfrac{1}{800}\left(x^3-75 x^2+675 x+30\ 375\right) & 5 \leq x \leq 55\end{cases}\)

\(f(x)\) is both smooth and continuous at \(x=5\).

The graph of  \(y=f(x)\)  is shown below, where \(x\) is the horizontal distance from the start of the ramp and \(y\) is the height of the ramp. All lengths are in centimetres.

  1. Find \(f^{\prime}(x)\) for \(0<x<55\).   (2 marks)

  2.   i. Find the coordinates of the point of inflection of \(f\).   (1 mark)

  3.  ii. Find the interval of \(x\) for which the gradient function of the ramp is strictly increasing.   (1 mark)

  4. iii. Find the interval of \(x\) for which the gradient function of the ramp is strictly decreasing.  (1 mark)

Jac and Jill decide to use two trapezoidal supports, each of width \(10 cm\). The first support has its left edge placed at \(x=5\) and the second support has its left edge placed at \(x=15\). Their cross-sections are shown in the graph below.

  1. Determine the value of the ratio of the area of the trapezoidal cross-sections to the exact area contained between \(f(x)\) and the \(x\)-axis between \(x=5\) and \(x=25\). Give your answer as a percentage, correct to one decimal place.   (3 marks)

  2. Referring to the gradient of the curve, explain why a trapezium rule approximation would be greater than the actual cross-sectional area for any interval \(x \in[p, q]\), where \(p \geq 25\).   (1 mark)

  3. Jac and Jill roll the toy car down the ramp and the car jumps off the end of the ramp. The path of the car is modelled by the function \(P\), where

      1. \(P(x)=\begin{cases}f(x) & 0 \leq x \leq 55 \\ g(x) & 55<x \leq a\end{cases}\)
  4. \(P\) is continuous and differentiable at \(x=55, g(x)=-\frac{1}{16} x^2+b x+c\), and \(x=a\) is where the car lands on the ground after the jump, such that \(P(a)=0\).
    1. Find the values of \(b\) and \(c\).   (2 marks)

    2. Determine the horizontal distance from the end of the ramp to where the car lands. Give your answer in centimetres, correct to two decimal places.   (1 mark)

Show Answers Only

a.    \(f^{\prime}(x)= \begin{cases}\displaystyle \ 0 & 0 < x<5 \\ \dfrac{1}{800}\left(3x^2-150 x+675 \right) & 5 \leq x < 55\end{cases}\)

b.i   \((25, 20)\)

b.ii  \([25, 55]\)

b.iii \([5, 25]\)

c.    \(98.1\%\)

d.    \(\text{In the interval }x\in [25, q]\text{ the curve is concave up.}\)

\(\therefore\ \text{The sloped edges of the trapeziums would be above }f(x)\)

\(\text{making the trapezium rule approximation greater than the}\)

\(\text{actual area.}\)

e.i   \(b=8.75, c=-283.4375\)

e.ii  \(34.10\ \text{cm}\)

Show Worked Solution

a.    \(\text{Using CAS: Define f(x) then}\)

\(\text{OR}\quad f^{\prime}(x)= \begin{cases}\displaystyle \ 0 & 0 < x<5 \\ \dfrac{1}{800}\left(3x^2-150 x+675 \right) & 5 \leq x < 55\end{cases}\)

b.i   \(\text{Using CAS: Find }f^”(x)\ \text{then solve }=0\)

\(\therefore\ \text{Point of inflection at }(25, 20)\)
  

b.ii  \(\text{Gradient function strictly increasing for }x\in [25, 55]\)

b.iii \(\text{Gradient function strictly decreasing for }x\in [5, 25]\)

c.    \(\text{Using CAS:}\)

\(\text{Area}\) \(=\dfrac{h}{2}\Big(f(5)+2f(15)+f(25)\Big)\)
  \(=\dfrac{10}{2}\Big(40+2\times 33.75+20\Big)=637.5\)

  

\(\therefore\ \text{Estimate : Exact}\) \(=637.5:650\)
  \(=\dfrac{637.5}{650}\times 100\)
  \(=98.0769\%\approx 98.1\%\)

    
d.    \(\text{In the interval }x\in [25, q]\text{ the curve is concave up.}\)

\(\therefore\ \text{The sloped edges of the trapeziums would be above }f(x)\)

\(\text{making the trapezium rule approximation greater than the}\)

\(\text{actual area.}\)
 

e.i   \(f(55)=\dfrac{35}{4}\ \text{(Using CAS)}\)

\(\text{and }f(55)=g(55)\rightarrow g(55)=\dfrac{35}{4}\)

\(\therefore -\dfrac{1}{16}\times 55^2+55b+c\) \(=\dfrac{35}{4}\)
\(c\) \(=\dfrac{35}{4}+\dfrac{3025}{16}-55b\)
\(c\) \(=\dfrac{3165}{16}-55b\quad (1)\)

  
\(g'(x)=-\dfrac{x}{8}+b\)

\(\text{and }g'(55)=f'(55)\)

\(\rightarrow\ f'(55)=\dfrac{1}{800}(3\times 55^2-150\times 55+675)=\dfrac{15}{8}\)

\(\rightarrow\ -\dfrac{55}{8}+b=\dfrac{15}{8}\)

\(\therefore b=\dfrac{35}{4}\quad (2)\)

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

\(c=\dfrac{3165}{16}-55\times\dfrac{35}{4}\)

\(c=-\dfrac{4535}{16}\)

\(\therefore\ b=\dfrac{35}{4}\ \text{or}\ 8.75 , c=-\dfrac{4535}{16}\ \text{or}\ -283.4375\)
 

e.ii  \(\text{Using CAS: Solve }g(x)=0|x>55\)

\(\text{Horizontal distance}=\sqrt{365}+70-55=34.104\dots\approx 34.10\ \text{cm (2 d.p.)}\)

PHYSICS, M7 2020 VCE 16 MC

The diagram below shows a plot of maximum kinetic energy, \(E_{\text{k max}}\),  versus frequency, \(f\), for various metals capable of emitting photoelectrons.
 

 

Which one of the following correctly ranks these metals in terms of their work function, from highest to lowest in numerical value?

  1. sodium, potassium, lithium, nickel
  2. nickel, potassium, sodium, lithium
  3. potassium, nickel, lithium, sodium
  4. lithium, sodium, potassium, nickel
Show Answers Only

\(D\)

Show Worked Solution
  • \(K_{\text{max}}=hf-\Phi \ \ \Rightarrow\ \  \Phi=hf-K_{\text{max}}\)
  • The threshold frequency occurs when \(K_{\text{max}}=0.\)
  • \(\Phi=hf_{\text{thresh}}\), hence  \(\Phi \propto f_{\text{thresh}}\).
  • The greater the threshold frequency, the greater the work function. Therefore, lithium has the greatest work function.

\(\Rightarrow D\)

Statistics, SPEC2 2022 VCAA 6

A company produces soft drinks in aluminium cans.

The company sources empty cans from an external supplier, who claims that the mass of aluminium in each can is normally distributed with a mean of 15 grams and a standard deviation of 0.25 grams.

A random sample of 64 empty cans was taken and the mean mass of the sample was found to be 14.94 grams.

Uncertain about the supplier's claim, the company will conduct a one-tailed test at the 5% level of significance. Assume that the standard deviation for the test is 0.25 grams.

  1. Write down suitable hypotheses \(H_0\) and \(H_1\) for this test.   (1 mark)

  2. Find the \(p\) value for the test, correct to three decimal places.   (1 mark)

  3. Does the mean mass of the random sample of 64 empty cans support the supplier's claim at the 5% level of significance for a one-tailed test? Justify your answer.   (1 mark)

  4. What is the smallest value of the mean mass of the sample of 64 empty cans for \(H_0\) not to be rejected? Give your answer correct to two decimal places.   (1 mark)

The equipment used to package the soft drink weighs each can after the can is filled. It is known from past experience that the masses of cans filled with the soft drink produced by the company are normally distributed with a mean of 406 grams and a standard deviation of 5 grams.

  1. What is the probability that the masses of two randomly selected cans of soft drink differ by no more than 3 grams? Give your answer correct to three decimal places.   (2 marks)

Show Answers Only

a.   \(H_0: \mu=15, \quad H_1: \mu<15\)

b.   \(p=0.027\)

c.    \(\text{Since \(\ p<0.05\), claim is not supported.}\)

d.   \(a=14.95\)

e.  \(\text{Pr}(-3<D<3)=0.329\)

Show Worked Solution

a.    \(H_0: \mu=15, \quad H_1: \mu<15\)
 

b.    \(\mu=15, \ \ \sigma=0.25\)

\(\bar{x}=14.94, \ \sigma_{\bar{x}}=\dfrac{0.25}{\sqrt{64}}=0.03125\)

\(\text{By CAS:}\)

\(p=\text{Pr}\left(\bar{X}<14.94 \mid \mu=15\right)=0.027\ \text {(3 d.p.)}\)
 

c.    \(\text{Since \(\ p<0.05\), claim is not supported.}\)

\(\text{Evidence is against \(H_0\)  at the \(5 \%\) level.}\)
 

d.    \(\text{Pr}\left(\bar{X}<a \mid \mu=15\right)>0.05\)

\(\text{Pr}\left(Z<\dfrac{a-15}{0.03125}\right)>0.05\)

\(\text{By CAS:}\ \ a=14.95\ \text{(2 d.p.)}\)
 

e.    \(\text{Let}\ \ M=\ \text{mass of one can}\)

\(M \sim N\left(406,5^2\right)\)

\(E\left(M_1\right)=E\left(M_2\right)=\mu=406\)

\(\text {Let}\ \ D=M_1-M_2\)

\(E(D)=406-406=0\)

\(\text{Var}(D)=1^2 \times \text{Var}\left(M_1\right)+(-1)^2 \times  \text{Var}\left(M_2\right)=50\)

\(\sigma_D=\sqrt{50}\)

\(D \sim N\left(0,(\sqrt{50})^2\right)\)

\(\text{By CAS: Pr\((-3<D<3)=0.329\) (3 d.p.) }\)

♦ Mean mark (e) 46%.

CHEMISTRY, M2 EQ-Bank 4

  1. Consider the compounds butyraldehyde \(\ce{(C4H8O)}\), lactic acid \(\ce{(C3H6O3)}\), and fructose \(\ce{(C6H12O6)}\).
  2. Identify which TWO of these compounds have the same empirical formula and justify your choice.    (2 marks)

  3. The empirical formula of a compound is \(\ce{C4H5O2}\) and its molar mass is determined to be 340.32 g mol\(^{-1}\).
  4. Calculate the molecular formula of this compound.     (3 marks)

Show Answers Only

a.    Determine the empirical formula of each compound.

Butyraldehyde \(\ce{(C4H8O)}\): Empirical formula = \(\ce{C4H8O}\)

Lactic acid \(\ce{(C3H6O3)}\): Empirical formula = \(\ce{CH2O}\)

Fructose \(\ce{(C6H12O6)}\): Empirical formula = \(\ce{CH2O}\)

  • Lactic acid and fructose have the same empirical formula of \(\ce{CH2O}\). 

b.    \(\ce{C16H20O8}\)

Show Worked Solution

a.    Determine the empirical formula of each compound.

Butyraldehyde \(\ce{(C4H8O)}\): Empirical formula = \(\ce{C4H8O}\)

Lactic acid \(\ce{(C3H6O3)}\): Empirical formula = \(\ce{CH2O}\)

Fructose \(\ce{(C6H12O6)}\): Empirical formula = \(\ce{CH2O}\)

  • Lactic acid and fructose have the same empirical formula of \(\ce{CH2O}\). 

b.    Calculate the molar mass of the empirical formula \(\ce{C4H5O2}\):

  \(\text{Molar mass}\ \ce{(C4H5O2)} = 4 \times 12.01 + 5 \times 1.008 + 2 \times 16.00 = 85.08\ \text{g/mol}\)
 

Ratio of the molar mass of the compound to the molar mass of the empirical formula:

  \(\text{Ratio} = \dfrac{\text{Molar mass}}{\text{Empirical formula mass}} = \dfrac{340.32\ \text{g mol}^{-1}}{85.08\ \text{g mol}^{-1}} =4\)
 

Multiply the subscripts in the empirical formula by 4 to get the molecular formula:

  \(\text{Molecular formula} = \ce{(C4H5O2)} \times 4 = \ce{C16H20O8}\)
 

  • Thus, the molecular formula of the compound is \(\ce{C16H20O8}\).

CHEMISTRY, M2 EQ-Bank 1

A propane tank contains 10.0 kg of propane \(\ce{(C3H8)}\). How many moles of propane are in the tank?   (2 marks)

Show Answers Only

 \(226.8\ \text{mol}\)

Show Worked Solution
  • Calculate the number of moles of propane (\(\ce{C3H8}\)):
  •    \(\ce{n(C3H8) = \dfrac{\text{m}}{\text{MM}} = \dfrac{10\,000\ \text{g}}{44.094 \, \text{g mol}^{-1}} = 226.79 \, mol}\)
  • The number of moles of propane in the tank is 226.8 mol.

Calculus, SPEC2 2022 VCAA 3

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

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

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

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

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

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

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

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

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

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

Show Answers Only

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

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

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

b.ii. 
           

c.    \(0.02\)

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

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

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

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

Show Worked Solution

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

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

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

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

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

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

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

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

♦♦♦ Mean mark 23%.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

CHEMISTRY, M1 EQ-Bank 13

You are given the task to separate the components of two mixtures: a saltwater solution and a mixture of sand and iron filings.

  1.  Suggest a suitable separation technique to extract the salt from the saltwater solution. Explain your reasoning based on the physical property involved.  (2 marks)
  2.  Identify the physical property that allow the mixture of sand and iron fillings to be separated and whether it is a homogeneous or heterogeneous mixture.  (1 marks)
  3. Describe one safety precaution that should be followed during the separation of the saltwater solution.  (2 marks)
Show Answers Only

a.   Separation technique:

  • Evaporation is a suitable technique because water has a much lower boiling point than salt.
  • By heating the solution, the water evaporates, leaving salt crystals behind.
  • Further distillation could be used to collect and condense the evaporated water.

b.   Physical property:

  • Iron fillings are magnetic and hence can be separated using a magnet.
  • The mixture is heterogeneous.

c.   Safety precaution:

  • For evaporation, wear heat-resistant gloves when handling hot equipment like the tripod or beaker to avoid burns to the hands.
Show Worked Solution

a.   Separation technique:

  • Evaporation is a suitable technique because water has a much lower boiling point than salt.
  • By heating the solution, the water evaporates, leaving salt crystals behind.
  • Further distillation could be used to collect and condense the evaporated water.

b.   Physical property:

  • Iron fillings are magnetic and hence can be separated using a magnet.
  • The mixture is heterogeneous.

c.   Safety precaution:

  • For evaporation, wear heat-resistant gloves when handling hot equipment like the tripod or beaker to avoid burns to the hands.

Functions, MET2 2022 VCAA 4

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

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

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

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

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

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

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

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

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

a.   `R` is the range.

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

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

c.   `f(x)+f(-x)`

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

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

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

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

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

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

`\frac{1}{k}(4)<1`   [Using CAS]

`:.\  k > 4`


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

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

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


♦♦♦♦ Mean mark (e.ii) 10%.

Calculus, SPEC2 2022 VCAA 1

Consider the family of functions \(f\) with rule  \(f(x)=\dfrac{x^2}{x-k}\), where \(k \in R \backslash\{0\}\).

  1. Write down the equations of the two asymptotes of the graph of \(f\) when \(k=1\).   (2 marks)

  2. Sketch the graph of  \(y=f(x)\)  for  \(k=1\)  on the set of axes below. Clearly label any turning points with their coordinates and label any asymptotes with their equations.   (3 marks)
     

  1.  i. Find, in terms of \(k\), the equations of the asymptotes of the graph of  \(f(x)=\dfrac{x^2}{x-k}\).   (1 mark)

  2. ii. Find the distance between the two turning points of the graph of  \(f(x)=\dfrac{x^2}{x-k}\) in terms of \(k\).   (2 marks)

  3. Now consider the functions \(h\) and \(g\), where  \(h(x)=x+3\)  and  \(g(x)=\abs{\dfrac{x^2}{x-1}}\).
  4. The region bounded by the curves of \(h\) and \(g\) is rotated about the \(x\)-axis.
    1. Write down the definite integral that can be used to find the volume of the resulting solid.   (2 marks)

    2. Hence, find the volume of this solid. Give your answer correct to two decimal places.   (1 mark)

Show Answers Only

a.  \(\text {Asymptotes: } x=1,\  y=x+1\)

b.   
       

c.i.   \(\text {Asymptotes: } x=k,\  y=x+k\)

c.ii.  \(\text {Distance }=2 \sqrt{5}|k|\)

d.i.  \(\displaystyle V=\pi \int_{\frac{-\sqrt{7}-1}{2}}^{\frac{\sqrt{7}-1}{2}}(x+3)^2-\left(\frac{x^2}{x-1}\right)^2 dx\)

d.ii.  \(V=51.42\ \text{u}^3 \)

Show Worked Solution

a.    \(\text {When } k=1 :\)

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

\(\text {Asymptotes: } x=1,\  y=x+1\)
 

b.    
       

 

c.i. \(f(x)=\dfrac{x^2}{x-k}=\dfrac{(x+k)(x-k)+k^2}{x-k}=x+k+\dfrac{k^2}{x-k}\)

\(\text {Using part a.}\)

\(\text {Asymptotes: } x=k,\  y=x+k\)
 

c.ii.  \(f^{\prime}(x)=1-\left(\dfrac{k}{x-k}\right)^2\)

\(\text {TP’s when } f^{\prime}(x)=0 \text { (by CAS):}\)

\(\Rightarrow(2 k, 4 k),(0,0)\)

\(\text {Distance }\displaystyle=\sqrt{(2 k-0)^2+(4 k-0)^2}=\sqrt{20 k^2}=2 \sqrt{5}|k|\)
 

d.i  \(\text {Solve for intersection of graphs (by CAS):}\)

\(\displaystyle x+3=\left|\frac{x^2}{x-1}\right|\)

\(\displaystyle \Rightarrow x=\frac{3}{2}, x=\frac{-1 \pm \sqrt{7}}{2}\)

\(\displaystyle V=\pi \int_{\frac{-\sqrt{7}-1}{2}}^{\frac{\sqrt{7}-1}{2}}(x+3)^2-\left(\frac{x^2}{x-1}\right)^2 dx\)
 

d.ii. \(V=51.42\ \text{u}^3 \text{ (by CAS) }\)

♦♦ Mean mark (d)(ii) 37%.

Vectors, SPEC2 2022 VCAA 13 MC

The acceleration of a body moving in a plane is given by  \(\underset{\sim}{\ddot{\text{r}}}(t)=\sin(t)\underset{\sim}{\text{i}}+2 \cos(t)\underset{\sim}{\text{j}}\), where  \(t \ge 0\).

Given that  \(\underset{\sim}{\dot{\text{r}}}(0)=2\underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}\), the velocity of the body at time \(t, \underset{\sim}{\dot{\text{r}}}(t)\), is given by

  1. \(-\cos (t) \underset{\sim}{\text{i}}+2 \sin (t) \underset{\sim}{\text{j}}\)
  2. \((3-\cos (t)) \underset{\sim}{\text{i}}+(2 \sin (t)+1) \underset{\sim}{\text{j}}\)
  3. \((1+\cos (t)) \underset{\sim}{\text{i}}+(2\sin (t)+1) \underset{\sim}{\text{j}}\)
  4. \((2+\sin (t)) \underset{\sim}{\text{i}}+(2\cos (t)-1) \underset{\sim}{\text{j}}\)
  5. \((1+\cos (t)) \underset{\sim}{\text{i}}+(1-2\sin (t)) \underset{\sim}{\text{j}}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\underset{\sim}{\ddot{r}}(t)=\sin (t) \underset{\sim}{i}+2 \cos (t) \underset{\sim}{j}\)

\(\underset{\sim}{\dot{r}}(t)=\displaystyle{\int} \ddot{r}(t) d t=-\cos (t) \underset{\sim}{i}+2 \sin (t) \underset{\sim}{j}+\underset{\sim}{c}\)

\(\text {When } t=0,\  \dot{r}(t)=2 \underset{\sim}{i}+\underset{\sim}{j}:\)

\begin{aligned}
2 \underset{\sim}{i}+\underset{\sim}{j} & =-\cos (0) \underset{\sim}{i}+0+\underset{\sim}{c} \\
\underset{\sim}{c} & =3\underset{\sim}{i}+\underset{\sim}{j}
\end{aligned}

\begin{aligned}
\therefore \dot{r}(t) & =-\cos(t)\underset{\sim}{i}+2 \sin (t) \underset{\sim}{j}+3 \underset{\sim}{i}+\underset{\sim}{j}\\
& =(3-\cos (t)) \underset{\sim}{i}+(2 \sin (t)+1) \underset{\sim}{j}
\end{aligned}

\(=>B\)

PHYSICS, M6 2020 VCE 6*

A single loop of wire moves into a uniform magnetic field \(B\) of strength 3.5 × 10\(^{-4}\) T over time \(t\) = 0.20 s from point \(\text{X}\) to point \(\text{Y}\) , as shown in the diagram below. The area \(A\) of the loop is 0.05 m\(^2\).
 

Determine the magnitude of the average induced EMF in the loop.   (2 marks)

Show Answers Only

\(8.8 \times 10^{-5}\ \text{V}\)

Show Worked Solution

\(\varepsilon=\dfrac{\Delta \Phi}{\Delta t}=\dfrac{B \Delta A}{\Delta t}=\dfrac{3.5 \times 10^{-4} \times 0.05}{0.2}=8.8 \times 10^{-5}\ \text{V}\)

PHYSICS, M5 2020 VCE 2*

Jupiter's moon Ganymede is its largest satellite.

Ganymede has a mass of 1.5 × 10\(^{23}\) kg and a radius of 2.6 × 10\(^6\) m.

Determine the magnitude of Ganymede's surface gravity?   (2 marks)

Show Answers Only

\(1.5\ \text{ms}^{-2}\)

Show Worked Solution
\(F\) \(=\dfrac{GMm}{r^2}\)  
\(mg\) \(=\dfrac{GMm}{r^2}\)  
\(g\) \(=\dfrac{GM}{r^2}\)  
  \(=\dfrac{6.67 \times 10^{-11} \times 1.5 \times 10^{23}}{(2.6 \times 10^6)^2}\)  
  \(=1.5\ \text{ms}^{-2}\)  

Functions, EXT1 F1 2023 MET1 7

Consider \(f:(-\infty, 1]\rightarrow R, f(x)=x^2-2x\). Part of the graph of  \(y=f(x)\)  is shown below.
 

  1. State the range of \(f\).   (1 mark)
  2. 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)
  3. Determine the equation of the domain for the inverse function  \(f^{-1}\).   (2 marks)
Show Answers Only

a.    \([-1, \infty)\)

b.   

c.    \(f^{-1}(x)=1-\sqrt{x+1}\)

\(\text{Domain}\ [-1, \infty)\)

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 \(f^{-1}(x)=1+\sqrt{x+1}\).

Functions, MET2 2022 VCAA 15 MC

The maximal domain of the function with rule `f(x)=\sqrt{x^2-2 x-3}` is given by

  1. `(-\infty, \infty)`
  2. `(-\infty,-3) \cup(1, \infty)`
  3. `(-1,3)`
  4. `[-3,1]`
  5. `(-\infty,-1] \cup[3, \infty)`
Show Answers Only

`E`

Show Worked Solution

`f(x)=\sqrt{x^2-2 x-3}`

`:. \ x^2-2 x-3 >= 0`

`:. \ (x  –  3)(x + 1)>=0`
  

So, `x <= -1` and `x >= 3`

`=>E`

 

Probability, MET2 2022 VCAA 10 MC

An organisation randomly surveyed 1000 Australian adults and found that 55% of those surveyed were happy with their level of physical activity.

An approximate 95% confidence interval for the percentage of Australian adults who were happy with their level of physical activity is closest to

  1. (4.1, 6.9)
  2. (50.9, 59.1)
  3. (52.4, 57.6)
  4. (51.9, 58.1)
  5. (45.2, 64.8)
Show Answers Only

`D`

Show Worked Solution

`\hat{P} = 0.55` , `n = 1000`, with approximate 95% confidence interval (`z=1.96`) for `\hat{P}`

Confidence Interval `= \left(\hat{p}-z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)`  
  `= \left(0.55-1.96 \sqrt{\frac{0.55 \times 0.45}{1000}}, 0.55+1.96 \sqrt{\frac{0.55 \times 0.45}{1000}}\right)`  
  `~~  \(0.519,0.581) \text { or }(51.9 \%, 58.1 \%)`  

 
`=>D`

Graphs, SPEC2 2022 VCAA 1 MC

For the interval  `\frac{1}{2} \le x \le3`, the graph of  `y=|2 x-1|-|x-3|`  is the same as the graph of

  1. `y=-x-2`
  2. `y=3x-4`
  3. `y=x+2`
  4. `y=3x+2`
  5. `y=x-4`
Show Answers Only

`B`

Show Worked Solution

`text{Consider the graph (CAS):}`

`text{Interval:}\ \ \frac{1}{2} \le x \le3`

`text{Endpoints:}\ \ (1/2 , -2\frac{1}{2}),\ \ (3 , 5)`

`m = frac{7\frac{1}{2}}{2\frac{1}{2}} = 3`

`y-y_1` `= m(x-x_1)`  
`y-5` `= 3(x-3)`  
`y` `= 3x-4`  

 
`=>B`

Vectors, EXT2 V1 2022 SPEC1 6

Find the cosine of the acute angle between the vectors  `underset~a=2underset~i-3underset~j+6underset~k`  and  `underset~b=underset~i+2underset~j+2underset~k`.   (2 marks)

Show Answers Only

`8/21`

Show Worked Solution
`cos \ theta` `=(underset~a*underset~b)/(|underset~a||underset~b|)`  
  `=(2-6+12)/(sqrt(4+9+36)\ sqrt(1+4+4))`  
  `=8/(7 xx 3)`  
  `=8/21`  

Vectors, SPEC1 2022 VCAA 6a

Find the cosine of the acute angle between the vectors  `underset~a=2underset~i-3underset~j+6underset~k`  and  `underset~b=underset~i+2underset~j+2underset~k`.   (2 marks)

Show Answers Only

`8/21`

Show Worked Solution
`cos \ theta` `=(underset~a*underset~b)/(|underset~a||underset~b|)`  
  `=(2-6+12)/(sqrt(4+9+36)\ sqrt(1+4+4))`  
  `=8/(7 xx 3)`  
  `=8/21`  

Complex Numbers, EXT2 N2 2022 SPEC1 1

Consider the equation  `p(z)=z^2 + 6iz-25`, `z ∈ C`.

  1. Express `p(z)` in the form  `p(z) = (z+ai)^2 + b`  where  `a`, ` b  ∈ R`.   (1 mark)
  2. Hence, or otherwise, find the solutions of the equation  `p(z) = 0`.   (2 marks)
Show Answers Only
  1. `p(z) = (z + 3i)^2-16`
  2. `z =  ± 4-3i`
Show Worked Solution
a.   `p(z)` `= (z + 3i)^2-(3i)^2-25`
    `=(z + 3i)^2 +9-25`
    `=(z + 3i)^2-16`

 

b.  
`(z + 3i)^2` `= 16`
  `z + 3i` `= ± 4`
  `z ` `= ± 4-3i`

Complex Numbers, SPEC1 2022 VCAA 1

Consider the equation  `p(z)=z^2 + 6iz - 25`, `z ∈ C`.

  1. Express `p(z)` in the form  `p(z) = (z+ai)^2 + b`  where  `a`, ` b  ∈ R`.   (1 mark)

  2. Hence, or otherwise, find the solutions of the equation  `p(z) = 0`.   (2 marks)

Show Answers Only
  1. `p(z) = (z + 3i)^2-16`
  2. `z =  ± 4-3i`
Show Worked Solution
a.   `p(z)` `= (z + 3i)^2-(3i)^2-25`
    `=(z + 3i)^2 +9-25`
    `=(z + 3i)^2-16`

 

b.  
`(z + 3i)^2` `= 16`
  `z + 3i` `= ± 4`
  `z ` `= ± 4-3i`

Statistics, SPEC2 2023 VCAA 6

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.

  1. Find a 95% confidence interval for the population mean (the mean mass of all adult male koalas in the forest). Give your values correct to two decimal places.  (1 mark)

  2. Sixty such random samples are taken and their confidence intervals are calculated.
  3. In how many of these confidence intervals would the actual mean mass of all adult male koalas in the forest be expected to lie?  (1 mark)

The ranger wants to decrease the width of the 95% confidence interval by 60% to get a better estimate of the population mean.

  1. How many adult male koalas should be sampled to achieve this?  (1 mark)

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.

  1. Write down the null hypothesis, \(H_0\), and the alternative hypothesis, \(H_1\), for the test.   (1 mark)

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.

  1.  i. Find the \(p\) value for the test correct to four decimal places.  (1 mark)

  2. ii. Draw a conclusion about the null hypothesis in part d. from the \(p\) value found above, giving a reason for your conclusion.  (1 mark)

  3. What is the critical sample mean (the smallest sample mean for \(H_0\) not to be rejected) in this test? Give your answer in kilograms correct to three decimal places.  (1 mark)

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%.

  1. What is the probability, correct to three decimal places, of the ranger making a type \(\text{II}\) error in the statistical test?  (1 mark)

Show Answers Only

a.    \((10.95,11.83)\)

b.    \(57\)

c.    \(n=125\)

d.    \(H_0: \mu=12, \quad H_1: \mu<12\)

e.i.  \(p=0.0057\)

e.ii. \(\text{Since } p<0.01 \text { : reject } H_0 \text {, favour } H_1\)

f.    \(\text {Critical sample mean } \bar{x} \approx 11.632\)

g.    \(\text{Pr}(\bar{x} \geqslant 11.63217 \mid \mu=11.4) \approx 0.071\)

Show Worked Solution

a.    \(\sigma_{\text{pop}}=1\)

\(\text{Sample:}\ \ n=20,\ \ \bar{x}=11.39,\ \ \sigma_{\text {sample }}=\dfrac{1}{\sqrt{20}}\)

\(\text{Find 95% C.I. (by CAS):}\)

\((10.95,11.83)\)
 

b.    \(\text{95% C.I. for 60 samples calculated}\)

\(\text{Number expected }(\mu \text{ within C.I.)}=0.95 \times 60=57\)
 

c.    \(\text {C.I.}=\left(11.39-1.96 \times \dfrac{1}{\sqrt{20}}, 11.39+1.96 \times \dfrac{1}{\sqrt{20}}\right)\)

\(\Rightarrow \text { Interval }=2 \times 1.96 \times \dfrac{1}{\sqrt{20}}\)

\(\text{Interval reduced by } 60\%\)

\(\Rightarrow \text{ New interval }=0.40 \times 2 \times 1.96 \times \dfrac{1}{\sqrt{20}}\)

\(\text{Solve for } n\) :

\(2 \times 1.96 \times \dfrac{1}{\sqrt{n}}=0.40 \times 2 \times 1.96 \times \dfrac{1}{\sqrt{20}}\)

\(\Rightarrow n=125\)
 

♦♦♦ Mean mark (c) 28%.

d.    \(H_0: \mu=12,\ \ H_1: \mu<12\)
 

e.i.  \(E(\bar{X})=\mu=12\)

 \(\bar{x}=11.6, \ \sigma(\bar{X})=\dfrac{1}{\sqrt{40}}\)

 \(p=\operatorname{Pr}(\bar{X} \leqslant 11.6)=0.0057\)

 

e.ii. \(\text{Since}\ \  p<0.01 \text {: reject } H_0 \text {, favour } H_1\)
 

f.    \(\text{Pr}(\bar{X} \leqslant a \mid \mu=12) \geqslant 0.01\) 

\(\text{Find } a \text{ (by CAS):}\)

\(\text{inv Norm}\left(0.01,12, \dfrac{1}{\sqrt{40}}\right) \ \Rightarrow \ a \geqslant 11.63217\)

\(\text {Critical sample mean}\ \ \bar{x} \approx 11.632\)
 

g.    \(\mu=11.4,\ \ \sigma_{\text{pop}}=1,\ \  n=40\)

\(\text{Pr}(\bar{X} \geqslant 11.63217 \mid \mu=11.4) \approx 0.071\)

♦♦ Mean mark (g) 39%.

PHYSICS, M7 2022 VCE 14*

Sam undertakes a photoelectric effect experiment using the apparatus shown in Figure 1. She uses a green filter.
 

   

Sam produces a graph of photocurrent, \(I\), in milliamperes, versus voltage, \(V\), in volts, as shown in Figure 2.
 

   

  1. Identify what point \(\text{P}\) represents on the graph in Figure 2.   (1 mark)
  1. Sam then significantly increases the intensity of the light.
  2. Sketch the resulting graph on Figure 3. The dashed line in Figure 14 represents the original data.   (2 marks)
     

   

  1. Sam replaces the green filter with a violet filter, keeping the light source at the increased intensity.
  2. Sketch the resulting graph on Figure 4. The dashed line in Figure 4 represents the original data.   (2 marks)
     

Show Answers Only

a.    Stopping voltage        

b.   

c.   

Show Worked Solution

a.    Stopping voltage
 

b.    Diagram will exhibit the following points:

  • Increasing the intensity of light will increase the photocurrent due to more photoelectrons being ejected from the metal.
  • However, it will not change the energy of the photoelectrons, hence the stopping voltage remains the same.
     

c.    Diagram will exhibit the following points:

  • Changing the colour from green to violet will increase the frequency of the light.
  • As \(E=hf\), the energy of the photoelectrons will increase, hence there needs to be a greater stopping voltage.
     

Calculus, MET2 2023 VCAA 1

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

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

a.    \((-1, 0), (0, 0), (2, 0)\)

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

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

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

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

c.iii. \(5.95\)

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

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

Show Worked Solution

a.    \((-1, 0), (0, 0), (2, 0)\)
  

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

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

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

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

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

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

ci    \(\text{Given }f(x)=g(x)\)

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

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

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

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

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

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

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

  

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

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

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

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

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

\(2ab+b^2=-2\ \ …(2)\)

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

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

  

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

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

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

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

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

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

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

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

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

 

PHYSICS, M5 2021 VCE 9-10 MC

Lucy is running horizontally at a speed of 6 m s\(^{-1}\) along a diving platform that is 8.0 m vertically above the water.

Lucy runs off the end of the diving platform and reaches the water below after time \(t\).

She lands feet first at a horizontal distance \(d\) from the end of the diving platform.
 

Question 9

Which one of the following expressions correctly gives the distance \(d\) ?

  1. 0.8\(t\)
  2. 6\(t\)
  3. 5\(t^2\)
  4. 6\(t\) + 5\(t^2\)


Question 10

Which one of the following is closest to the time taken, \(t\), for Lucy to reach the water below?

  1. 0.8 s
  2. 1.1 s
  3. 1.3 s
  4. 1.6 s
Show Answers Only

\(\text{Question 9:} \ B\)

\(\text{Question 10:} \ C\)

Show Worked Solution

Question 9

  • The horizontal displacement formula for projectile motion:  \(s=ut\)
  • Horizontal distance: \(d=6t\)

\(\Rightarrow B\)
 

Question 10

Find time of flight \((t):\)

\(s\) \(=ut+\dfrac{1}{2}at^2\)  
\(s\) \(=\dfrac{1}{2}at^2\ \ \ (u=0)\)  
\(t\) \(=\sqrt{\dfrac{2s}{a}}=\sqrt{\dfrac{2 \times 8}{9.8}}\approx 1.3\ \text{s}\)  

 

\(\Rightarrow C\)

Functions, 2ADV F2 2023 MET1 3

  1. 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)
     


  2. Find the values of \(x\) for which \(f(x)\leq1\).   (1 mark)
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).

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. The two diagrams below show a front view and top view of the car.
 

  1. Calculate the magnitude of the net force acting on the racing car and driver as they go around the corner.   (2 marks)
  1. On the "Top view" diagram, draw the direction of the net force acting on the racing car using an arrow.   (1 mark)
  1. Explain why the racing car needs a net horizontal force to travel around the corner and state what exerts this horizontal force.   (2 marks)
Show Answers Only

a.    \(1.6 \times 10^4\ \text{N}\)

b.    
       

c.   Net horizontal force:

  • 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 Worked Solution

a.    \(F=\dfrac{mv^2}{r}=\dfrac{800 \times (40)^2}{80}=1.6 \times 10^4\ \text{N}\)

b.   
       

c.   Net horizontal force:

  • 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.
♦ Mean mark 42%.

PHYSICS, M6 2022 VCE 6

The diagram shows a simple alternator consisting of a rectangular coil of area 0.060 m\(^{2}\) and 200 turns, rotating in a uniform magnetic field. The magnetic flux through the coil in the vertical position shown in the diagram is 1.2 × 10\(^{-3}\) Wb.
 

  1. Calculate the strength of the magnetic field. Show your working.  (2 marks)

  1. The rectangular coil rotates at a frequency of 2.5 Hz.
  2. Calculate the average induced EMF produced in the first quarter of a turn. Begin the quarter with the coil in the vertical position shown in the diagram.  (3 marks)

Show Answers Only

a.    \(0.02\ \text{T}\)

b.    \(2.4\ \text{V}\)

Show Worked Solution
a.     \(\Phi\) \(=BA\)
  \(B\) \(=\dfrac{\Phi}{A}=\dfrac{1.2 \times 10^{-3}}{0.060}=0.02\ \text{T}\)

 
b.    \(\text{The time for 1 complete rotation:}\)

\(T=\dfrac{1}{f}=\dfrac{1}{2.5}=0.4\ \text{s}\)

\(\Rightarrow \text{Time for a quarter turn}\ =0.1\ \text{s}\)

\(\varepsilon=N\dfrac{\Delta \Phi}{t}=200 \times  \dfrac{1.2 \times 10^{-3}}{0.1}=2.4\ \text{V}\)

PHYSICS, M6 2022 VCE 5*

A wind generator provides power to a factory located 2.00 km away, as shown in the diagram.

When there is a moderate wind blowing steadily, the generator produces a voltage of 415 V and a current of 100 A.

The total resistance of the transmission wires between the wind generator and the factory is 2.00 \(\Omega\).
 

  1. Calculate the power, in kilowatts, produced by the wind generator when there is a moderate wind blowing steadily.   (1 mark)

To operate correctly, the factory's machinery requires a power supply of 40 kW.

  1. Determine whether the energy supply system, as shown, will be able to supply power to the factory when the moderate wind is blowing steadily. Justify your answer with calculations.   (3 marks)

  1. The factory's owner decides to limit transmission energy loss by installing two transformers: a step-up transformer with a turns ratio of 1:10 at the wind generator and a step-down transformer with a turns ratio of 10:1 at the factory. Each transformer can be considered ideal.

    With the installation of the transformers, determine the power, in kilowatts, now supplied to the factory.   (3 marks)

Show Answers Only

a.    \(41.5\ \text{kW}\)

b.    \(P_{\text{loss}}=I^2R=100^2 \times 2=20000\ \text{W}=20\ \text{kW}\)

\(\Rightarrow\ \text{Net power supplied}\ =41.5-20=21.5\ \text{kW}\ < 40\ \text{kW}\)

\(\therefore \ \text{The power supply will not be enough to power the factory.}\)

c.    \(41.3\ \text{kW}\)

Show Worked Solution

a.    \(P=VI=415 \times 100 = 41\ 500\ \text{W}=41.5\ \text{kW}\)
 

b.    \(P_{\text{loss}}=I^2R=100^2 \times 2=20000\ \text{W}=20\ \text{kW}\)

\(\Rightarrow\ \text{Net power supplied}\ =41.5-20=21.5\ \text{kW}\ < 40\ \text{kW}\)

\(\therefore \ \text{The power supply will not be enough to power the factory.}\)
 

c.     \(\dfrac{I_s}{I_p}\) \(=\dfrac{N_p}{N_s}\)
  \(I_s\) \(=\dfrac{N_p}{N_s} \times I_p=\dfrac{1}{10} \times 100=10\ \text{A}\)

 

\(P_{\text{loss}} \text{(new)} =I^2R=10^2 \times 2=200\ \text{W}=0.2\ \text{kW}\)

\(\therefore\ \text{Net power supplied}\ =41.5-0.2=41.3\ \text{kW}\)

PHYSICS, M6 2022 VCE 1

The diagram shows four positions (1, 2, 3 and 4) of the coil of a single-turn, simple DC motor. The coil is turning in a uniform magnetic field that is parallel to the plane of the coil when the coil is in Position 1, as shown.

When the motor is operating, the coil rotates about the axis through the middle of sides \(L M\) and \(N K\) in the direction indicated. The coil is attached to a commutator. Current for the motor is passed to the commutator by brushes that are not shown in the diagram.
 

  1. When the coil is in Position 1, in which direction is the current flowing in the side \(K L-\) from \(K\) to \(L\) or from \(L\) to \(K\)? Justify your answer.   (2 marks)

  1. When the coil is in Position 3, in which direction is the current flowing in the side \(KL-\) from \(K\) to \(L\) or
    from \(L\) to \(K\)?       (1 marks)

  1. The side \(K L\) of the coil has a length of 0.10 m and experiences a magnetic force of 0.15 N due to the magnetic field, which has a magnitude of 0.5 T.

    Calculate the magnitude of the current in the coil.   (2 marks)

Show Answers Only

a.    \(K\) to \(L\)

b.    \(L\) to \(K\)

c.    \(3\ \text{A}\)

Show Worked Solution

a.    Using the right hand rule:

  • Palm faces up (force), fingers to the right (magnetic field), and the thumb faces out of the page.
  • Therefore the current must run from \(K\) to \(L\).

b.    Using the right hand rule:

  • Palm now faces down the page (force), fingers still to the right.
  • The current will run from \(L\) to \(K\) (thumb).
c.    \(F\) \(=lIB\)
  \(I\) \(=\dfrac{F}{lB}=\dfrac{0.15}{0.1 \times 0.5}=3\ \text{A}\)

PHYSICS, M7 2022 VCE 18 MC

Which one of the following is an example of an inertial frame of reference?

  1. a bus travelling at constant velocity
  2. an express train that is accelerating
  3. a car turning a corner at a constant speed
  4. a roller-coaster speeding up while heading down a slope
Show Answers Only

\(A\)

Show Worked Solution
  • \(A\) is the only frame of reference where no acceleration is imposed on the object.
  • It is therefore an inertial frame of reference.

\(\Rightarrow A\)

PHYSICS, M7 2022 VCE 16 MC

Which one of the following phenomena best demonstrates that light waves are transverse?

  1. polarisation
  2. interference
  3. dispersion
  4. diffraction
Show Answers Only

\(A\)

Show Worked Solution
  • Polarisation requires the transfer of energy through a wave to be perpendicular to the direction of oscillation of the wave.
  • As this is a property of a transverse wave, light waves must be traverse.

\(\Rightarrow A\)

PHYSICS, M7 2022 VCE 15 MC

Which one of the following best provides evidence of light behaving as a particle?

  1. photoelectric effect
  2. white light passing through a prism
  3. diffraction of light through a single slit
  4. interference of light passing through a double slit
Show Answers Only

\(A\)

Show Worked Solution
  • During the photoelectric effect, light transfers energy in small discrete ‘quanta’ which demonstrates light behaving as a particle.

\(\Rightarrow A\)

PHYSICS, M8 2022 VCE 3 MC

Particles emitted from a radioactive source travel through a magnetic field, \(B_{\text {in }}\), directed into the page, as shown schematically in the diagram below.

Three particles, \(\text{K, L}\) and \(\text{M}\), follow the paths indicated by the arrows.
 

Which of the following correctly identifies the charges on particles \(\text{K, L}\) and \(\text{M}\)?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|c|}
\hline \rule{0pt}{1.5ex}\textbf{K} \rule[-0.5ex]{0pt}{0pt}&\textbf{L} & \textbf{M} \\
\hline \rule{0pt}{2.5ex}\text{positive} \rule[-1ex]{0pt}{0pt}& \text{no charge} & \text{negative} \\
\hline \rule{0pt}{2.5ex}\text{positive} \rule[-1ex]{0pt}{0pt}& \text{negative} & \text{negative} \\
\hline \rule{0pt}{2.5ex}\text{negative} \rule[-1ex]{0pt}{0pt}& \text{no charge} & \text{positive} \\
\hline \rule{0pt}{2.5ex}\text{no charge} \rule[-1ex]{0pt}{0pt}& \text{no charge} & \text{no charge} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution
  • \(\text{L}\) must have no charge as it travels through the magnetic field with no deflection.
  • Using the right-hand rule, the force on a positive charge would be up the page, hence \(\text{K}\) must be the positive charge.

\(\Rightarrow A\)

PHYSICS, M6 2022 VCE 1 MC

A single loop of wire carries a current, \(I\), as shown in the diagram below.
 

Which one of the following best describes the direction of the magnetic field at the centre of the circle, \(\text{C}\), which is produced by the current carrying wire?

  1. to the left
  2. to the right
  3. into the page
  4. out of the page
Show Answers Only

\(C\)

Show Worked Solution
  • Right hand grip rule: fingers curl in a clockwise direction and the thumb points into the page.

\(\Rightarrow C\)

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.

  1. The above logistic differential equation can be expressed as
  2. \(\displaystyle \int \frac{A}{P}+\dfrac{B}{1-\dfrac{P}{1000}} d P=\int dt \text {, where } A, B \in R .\)
  3. Find the values of \(A\) and \(B\).  (1 mark)

One form of the solution for \(P\) is  \(P=\dfrac{1000}{1+D e^{-t}}\ \),  where \(D\) is a real constant.

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

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.

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

  2. Find the value of \(Q\) when \(t=6\).
  3. Give your answer correct to the nearest integer.  (1 mark)

  4.  i. Given that  \(\dfrac{dQ}{dt}=\dfrac{11}{10} Q\left(1-\dfrac{Q}{1000}\right)\),  express  \(\dfrac{d^2 Q}{d t^2}\)  in terms of \(Q\).  (1 mark)

  5. ii. Hence or otherwise, find the size of the fish population in pond 2 and the value of \(t\) when the rate of growth of the population is a maximum. Give your answer for \(t\) correct to the nearest year.  (2 marks)

  6. Sketch the graph of \(Q\) versus \(t\) on the set of axes below. Label any axis intercepts and any asymptotes with their equations.  (2 marks)
     

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

\(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)-0.055Q\)

  1. Find the maximum number of fish that could be supported in pond 2 in this situation.  (1 mark)

Show Answers Only

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.  \(Q=950\)

Show Worked Solution

a.    \(\dfrac{dP}{dt}=P\left(1-\dfrac{P}{1000}\right) \ \Rightarrow \ \dfrac{d t}{d P}=\dfrac{1}{P}\left(\dfrac{1}{1-\frac{P}{1000}}\right)\)

\(\text{Expand (by CAS):}\)

\(\dfrac{1}{P}\left(\dfrac{1}{1-\frac{P}{1000}}\right)=\dfrac{1}{P}-\dfrac{1}{(P-1000)}=\dfrac{1}{P}+\dfrac{1}{1000}\left(\dfrac{1}{1-\frac{P}{1000}}\right)\)

\(A=1, \quad B=\dfrac{1}{1000}\)
 

♦ Mean mark (a) 48%.

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}
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%.

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.  \(\text{Solve for } Q \ \text{(by CAS):}\)

\(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)-0.055\,Q=0\)

\(\Rightarrow Q=950\)

♦ Mean mark (g) 40%.

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.

  1.  i. Write down the definite integral, in terms of \(x\), for the volume of this solid of revolution.  (1 mark)

  2. ii. Find the volume of the solid of revolution.  (1 mark)

  3.  i. Express the curved surface area of the solid in the form  \(\pi \displaystyle \int_a^b \sqrt{A x-B}\, d x\), where \(a, b, A, B\) are all positive integers.  (2 marks)

  4. ii. Hence or otherwise, find the curved surface area of the solid correct to three decimal places.  (1 mark)

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.

  1. Find the efficiency ratio of the solid of revolution correct to two decimal places.  (2 marks)

  2. Another solid of revolution is formed by rotating the curve given by  \(y^2=x-1\) about the \(x\)-axis for  \(2 \leq x \leq k\), where \(k \in R\). This solid has a volume of \(24 \pi\).
  3. Find the efficiency ratio for this solid, giving your answer correct to two decimal places.  (3 marks)

Show Answers Only

a.i. \(\displaystyle V=\pi \int y^2 d x=\pi \int_2^5(x-1) dx\)
 

a.ii. \(\text{Evaluating integral (by calc):}\)

\(V=\dfrac{15 \pi}{2}\ \text{u}^3\)
 

b.i.  \(\displaystyle y=\sqrt{x-1} \ \Rightarrow \ \frac{d y}{d x}=\frac{1}{2} \cdot \frac{1}{\sqrt{x-1}} \ \Rightarrow \ \frac{d^2 y}{dx^2}=\frac{1}{4(x-1)}\)

\(\ \ \begin{aligned} \displaystyle \int_2^5 2 \pi \sqrt{x-1} \ \sqrt{1+\frac{1}{4(x-1)}} \ d x & =\int_2^5 2 \pi \sqrt{x-1+\frac{1}{4}}\, d x \\ & =\pi \int_2^5 \sqrt{4 x-3}\, d x\end{aligned}\)

 
b.ii. \(\text{Evaluate integral in b.i.}\)

\(\text {S.A.}=30.847\ \text{u}^2\)
 

c.    \(y=\sqrt{x-1}\)

\(\text{At}\ \ x=2\ \ \Rightarrow y=1\)

\(\text{At}\ \ x=5\ \ \Rightarrow y=2\)

\begin{aligned}
\text{Total S.A. } &=30.847+\pi(1)^2+\pi(2)^2 \\
&=46.5545\ \text{u}^2 \\
\end{aligned}

\(\text {Efficiency ratio}=\dfrac{46.5545}{\frac{15 \pi}{2}}=1.98\ \text{(2 d.p.)}\)
 

d.   \(V=\displaystyle \pi \int_2^k(x-1) d x=24 \pi\)

\(\text{Solve (by calc):}\)

\(\dfrac{\pi k(k-2)}{2}=24 \pi \ \Rightarrow \ k=8 \)

\(\text {S.A.}=\pi \displaystyle{\int_2^8} \sqrt{4 x-3}\, d x=75.9163 \ldots\)

\(\text{Total S.A.}=75.9163+\pi(1+7)=101.049 \)

\(\text{Efficiency ratio}=\dfrac{101.049}{24 \pi}= 1.34\ \text{(2 d.p.)}\)

Show Worked Solution

a.i. \(\displaystyle V=\pi \int y^2 d x=\pi \int_2^5(x-1) dx\)
 

a.ii. \(\text{Evaluating integral (by calc):}\)

\(V=\dfrac{15 \pi}{2}\ \text{u}^3\)
 

b.i.  \(\displaystyle y=\sqrt{x-1} \ \Rightarrow \ \frac{d y}{d x}=\frac{1}{2} \cdot \frac{1}{\sqrt{x-1}} \ \Rightarrow \ \frac{d^2 y}{dx^2}=\frac{1}{4(x-1)}\)

\(\ \ \begin{aligned} \displaystyle \int_2^5 2 \pi \sqrt{x-1} \ \sqrt{1+\frac{1}{4(x-1)}} \ d x & =\int_2^5 2 \pi \sqrt{x-1+\frac{1}{4}}\, d x \\ & =\pi \int_2^5 \sqrt{4 x-3}\, d x\end{aligned}\)

 
b.ii. \(\text{Evaluate integral in b.i.}\)

\(\text {S.A.}=30.847\ \text{u}^2\)
 

c.    \(y=\sqrt{x-1}\)

\(\text{At}\ \ x=2\ \ \Rightarrow y=1\)

\(\text{At}\ \ x=5\ \ \Rightarrow y=2\)

\begin{aligned}
\text{Total S.A. } &=30.847+\pi(1)^2+\pi(2)^2 \\
&=46.5545\ \text{u}^2 \\
\end{aligned}

\(\text {Efficiency ratio}=\dfrac{46.5545}{\frac{15 \pi}{2}}=1.98\ \text{(2 d.p.)}\)
 

d.   \(V=\displaystyle \pi \int_2^k(x-1) d x=24 \pi\)

\(\text{Solve (by calc):}\)

\(\dfrac{\pi k(k-2)}{2}=24 \pi \ \Rightarrow \ k=8 \)

\(\text {S.A.}=\pi \displaystyle{\int_2^8} \sqrt{4 x-3}\, d x=75.9163 \ldots\)

\(\text{Total S.A.}=75.9163+\pi(1+7)=101.049 \)

\(\text{Efficiency ratio}=\dfrac{101.049}{24 \pi}= 1.34\ \text{(2 d.p.)}\)

Complex Numbers, SPEC2 2023 VCAA 2

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

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

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

     

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

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

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

use De Moivre's theorem to show that

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

Show Answers Only

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

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

b.  \(\text{Roots of}\ \ z^7-1=0:\) 

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

c.   
          

d.i.   
          

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

\(\arg(z-1)=\dfrac{9\pi}{14} \)
 

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

 

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

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

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

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

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

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

Show Worked Solution

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

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

b.  \(\text{Roots of}\ \ z^7-1=0:\) 

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

c.   
          

d.i.   
          

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

\(\arg(z-1)=\dfrac{9\pi}{14} \)
 

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

 

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

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

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

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

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

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

Probability, MET2 2023 VCAA 4

A manufacturer produces tennis balls.

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

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

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

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

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

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

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

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

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

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

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

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

a.    \(0.1587\)

b.    \(d\approx6.83\)

c.    \(0.9938\)

d.    \(0.9998\)

e.    \(0,8960\)

f.    \(0.06\)

g.    \(90\%\)

h.    \(0.1345\)

i.    \(3(\pi^2+4)\)

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

Show Worked Solution

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

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

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

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

b.    \(\Pr(D<d)=0.90\)  

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

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

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

 

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

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

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

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

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

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

\(X\sim \text{Bi}(4, 0.999379 …)\)

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

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

 

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

  

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

 

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

♦♦♦ Mean mark (j) 10%.