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Probability, MET1 2023 VCAA SM-Bank 7

The duration of telemarketing calls to mobile phone users is a continuous random variable \(T\) minutes, with probability density function

\(f(t)= \begin{cases} \dfrac{2}{5} e^{-\frac{2}{5} t} & t \geq 0 \\ \ 0 & \text {elsewhere }\end{cases}\)

Find the value of \(k\) such that 90% of telemarketing calls last less than \(k\) minutes. Express your answer in the form  \(\dfrac{a}{b} \,\log _e(c)\),  where \(a, b\) and \(c\) are positive integers.   (3 marks)

Show Answers Only

\(k=\dfrac{5}{2}\log_{e}{10}\)

Show Worked Solution

\(\text{Pr}(T<k)=\displaystyle\int_{0}^{k}\dfrac{2}{5} e^{-\frac{2}{5} t}\,dt=0.9\)

\(\dfrac{2}{5}\Bigg[-\dfrac{5}{2}e^{-\frac{2t}{5}}\displaystyle\Bigg]_{0}^{k}\) \(=0.9\)
\(-e^{-\frac{2k}{5}}+1\) \(=0.9\)
\(e^{-\frac{2k}{5}}\) \(=0.1=\dfrac{1}{10}\)
\(e^{\frac{2k}{5}}\) \(=10\)
\(\dfrac{2k}{5}\) \(=\log_{3}{10}\)
\(\therefore\ k\) \(=\dfrac{5}{2}\log_{3}{10}\)

Calculus, MET1 2023 VCAA SM-Bank 5

Let  \(f: R \rightarrow R\),  where  \(f(x)=2-x^2\).

  1. Calculate the average rate of change of \(f\) between \(x=-1\) and \(x=1\).  (1 mark)

  2. Calculate the average value of \(f\) between \(x=-1\) and \(x=1\).  (2 marks)

  3. Four trapeziums of equal width are used to approximate the area between the functions  \(f(x)=2-x^2\)  and the \(x\)-axis from \(x=-1\) to \(x=1\).
  4. The heights of the left and right edges of each trapezium are the values of \(y=f(x)\), as shown in the graph below.

  1. Find the total area of the four trapeziums.  (2 marks)

Show Answers Only

a.    \(0\)

b.    \(\dfrac{5}{3}\)

c.    \(\dfrac{13}{4}\)

Show Worked Solution
a.     \(\text{Average rate of change}\) \(=\dfrac{f(1)-f(-1)}{1-(-1)}\)
    \(\dfrac{1-1}{2}\)
    \(=0\)

 

b.    \(\text{Avg value}\) \(=\dfrac{1}{1-(-1)}\displaystyle\int_{-1}^{1} \Big(2-x^2\Big)\,dx\)
    \(=\dfrac{1}{2}\displaystyle\Bigg[2x-\dfrac{x^3}{3}\displaystyle\Bigg]_{-1}^{1}\)
    \(=\dfrac{1}{2}\displaystyle\Bigg[\Bigg(2.(1)-\dfrac{(1)^3}{3}\Bigg)-\Bigg(2.(-1)-\dfrac{(-1)^3}{3}\Bigg)\Bigg]\)
    \(=\dfrac{1}{2}\times\dfrac{10}{3}=\dfrac{5}{3}\)

 

c.     \(\text{Total Area}\) \(=2\times\ \text{Area from 0 to 1}\)
    \(=2\times \dfrac{h}{2}\Big(f(0)+2.f(0.5)+f(1)\Big)\quad \text{where }h=\dfrac{1}{2}\)
    \(=\dfrac{1}{2}\Big(2+2\times\dfrac{7}{4}+1\Big)=\dfrac{13}{4}\)

Functions, MET1 EQ-Bank 4

Consider the simultaneous equations below, where \(a\) and \(b\) are real constants.

\begin{aligned}
& (a+3) x+9 y=3 b\\
& 2 x+a y=5
\end{aligned}

Find the values of \(a\) and \(b\) for which the simultaneous equations have no solutions.   (4 marks)

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\(a=-6\ \text{and}\ b\neq -\dfrac{5}{2}\ \ \text{OR}\ \ a=3\ \text{and}\ b\neq 5\)

Show Worked Solution

(\(a+3)x+9y\) \(=3b\)  
\(y\) \(=-\dfrac{a+3}{9}x+\dfrac{b}{3}\) \(\dots (1)\)
\(\rightarrow \ m_1\) \(=-\dfrac{a+3}{9}, \ c_1=\dfrac{b}{3}\)  

 

\(2x+ay\) \(=5\)  
\(y\) \(=-\dfrac{2}{a}x+\dfrac{5}{a}\) \(\dots (2)\)
\(\rightarrow \ m_2\) \(=-\dfrac{2}{a}, \ c_2=\dfrac{5}{a}\)  

  
\(\text{No solution if  }m_1=m_2,\ \text{and}\ c_1\neq c_2.\)

\(\rightarrow\ -\dfrac{a+3}{9}=-\dfrac{2}{a}\ \text{and }\dfrac{b}{3}\neq \dfrac{5}{a}\rightarrow\ b\neq \dfrac{15}{a}\)

   

\(-\dfrac{a+3}{9}=-\dfrac{2}{a}\)

\(a^2+3a-18\) \(=0\)  
\((a+6)(a-3)\) \(=0\)  
\(\therefore a=-6\ \) \(\ \text{or}\ \) \(\ a=3\)

  

\(\text{When }\ a=-6,\  b\neq \dfrac{15}{-6}\rightarrow\ b\neq -\dfrac{5}{2}\)

\(\text{OR}\)

\(\text{When }\ a=3,\  b\neq \dfrac{15}{3}\rightarrow\ b\neq 5\)

CHEMISTRY, M6 2008 HSC 14 MC

20 mL 0.08 mol L\(^{-1}\ \ce{HCl}\) is mixed with 30 mL of 0.05 mol L\(^{-1}\ \ce{NaOH}\).

What is the pH of the resultant solution?

  1. 1.1
  2. 2.7
  3. 4.0
  4. 7.0
Show Answers Only

\(B\)

Show Worked Solution

\(\ce{n(HCl)_{excess} = 0.0016-0.0015=0.0001 mol}\)

\[\ce{[HCl] = \frac{0.0001}{0.020+0.030} = 0.002 mol L^{-1}}\]

\(\ce{[HCl] = [H+]}\)

\(\ce{pH = -log(0.002) = 2.7}\)

\(\Rightarrow B\)

PHYSICS, M8 2020 VCE 17

Figure 17 shows the emission spectrum for helium gas.
 

   

  1. Which spectral line indicates the photon with the lowest energy?  (1 mark)
  1. Calculate the frequency of the photon emitted at the 588 nm line. Show your working.  (2 marks)
  1. Explain why only certain wavelengths and, therefore, certain energies are present in the helium spectrum.  (2 marks)

Show Answers Only

a.    \(668\ \text{nm}\)

b.    \(5.1 \times 10^{14}\ \text{Hz}\)

c.    → Electrons in the helium atom exist within electron shells.

→ Electrons in each shell have a fixed amount of energy. 

→ he electrons can absorb certain amounts of energy and move up into the next energy shell which is equal to the difference in energy between the shells.

→ When electrons fall back to their original energy state, they emit a photon with energy equal to the difference in energy between the shells.

→ As these are always fixed energy levels, only certain wavelengths and energies are emitted and therefore present in the helium spectrum.

Show Worked Solution

a.    \(E=\dfrac{hc}{\lambda}\),  therefore \(E \propto \dfrac{1}{\lambda}\)

→ The photon with the lowest energy will have the highest wavelength.

→ Lowest energy spectral line = 668 nm
 

b.    Convert: 588 nm = 588 × 10\(^{-9}\) m

\(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^8}{588 \times 10^{-9}}=5.1 \times 10^{14}\ \text{Hz}\)
 

c.    → Electrons in the helium atom exist within electron shells.

→ Electrons in each shell have a fixed amount of energy. 

→ he electrons can absorb certain amounts of energy and move up into the next energy shell which is equal to the difference in energy between the shells.

→ When electrons fall back to their original energy state, they emit a photon with energy equal to the difference in energy between the shells.

→ As these are always fixed energy levels, only certain wavelengths and energies are emitted and therefore present in the helium spectrum.

♦♦ Mean mark (c) 34%.

PHYSICS, M8 2020 VCE 16

A beam of electrons travelling at 1.72 × 10\(^5\) m s\(^{-1}\) illuminates a crystal, producing a diffraction pattern as shown in Figure 16. Ignore relativistic effects.
 

  1. Calculate the kinetic energy of one of the electrons. Show your working.  (2 marks)

  2. The electron beam is now replaced by an X-ray beam. The resulting diffraction pattern has the same spacing as that produced by the electron beam.

    Calculate the energy of one X-ray photon. Show your working.  (3 marks)

Show Answers Only

a.    \(0.084\ \text{eV}\)

b.    \(293\ \text{eV}\)

Show Worked Solution

a.     \(KE\) \(=\dfrac{1}{2}mv^2\)
    \(=\dfrac{1}{2} \times 9.109 \times 10^{-31} \times (1.72 \times 10^5)^2\)
    \(=1.3474\times 10^{-20}\ \text{J}\)
    \(=\dfrac{1.3474 \times 10^{-20}}{1.602 \times 10^{-19}}\)
    \(=0.084\ \text{eV}\)

 

b.     \(\lambda_e\) \(=\dfrac{h}{mv}\)
    \(=\dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 1.72 \times 10^5}\)
    \(=4.23 \times 10^{-9}\ \text{m}\)
    \(=\lambda_{\text{x-ray}}\)

 

  \(E_{\text{x-ray}}\) \(=\dfrac{hc}{\lambda}\)
    \(=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{4.23 \times 10^{-9}}\)
    \(=4.7 \times 10^{-17}\ \text{J}\)
    \(=\dfrac{4.7 \times 10^{-17}}{1.602 \times 10^{-19}}\)
    \(=293\ \text{eV}\)
♦ Mean mark (b) 41%.

PHYSICS, M7 2020 VCE 15

The metal surface in a photoelectric cell is exposed to light of a single frequency and intensity in the apparatus shown in Figure 14.

The voltage of the battery can be varied in value and reversed in direction.
  

  1. A graph of photocurrent versus voltage for one particular experiment is shown in Figure 15.
  2. On Figure 15, draw the trace that would result for another experiment using light of the same frequency but with triple the intensity.  (2 marks)
     

  1. What is a name given to the point labelled \(\text{A}\) on Figure 15?  (1 mark)

  2. Why does the photocurrent fall to zero at the point labelled \(\text{A}\) on Figure 15 ?  (1 mark)

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a.   
         

b.    The stopping voltage.

c.    → The energy of the stopping voltage is equal to the most energetic photoelectrons. 

→ Therefore, the stopping voltage will have enough energy to turn back/stop all of the photoelectrons.

Show Worked Solution

a.    The photocurrent will also be tripled.
 

b.    The stopping voltage.
 

c.    → The energy of the stopping voltage is equal to the most energetic photoelectrons. 

→ Therefore, the stopping voltage will have enough energy to turn back/stop all of the photoelectrons.

♦♦ Mean mark (c) 32%.
COMMENT: The work function of the metal had no relevance to this question.

PHYSICS, M7 2020 VCE 14

The diagram below shows a representation of an electromagnetic wave.

Correctly label Figure 13 using the following symbols.  (3 marks)

 

Show Answers Only

Show Worked Solution

 
→ The wavelength is from the first peak on the electric wave to the second peak on the electric wave (not from an electric wave peak to magnetic wave peak).

PHYSICS, M5 2020 VCE 8

Figure 8 shows a small ball of mass 1.8 kg travelling in a horizontal circular path at a constant speed while suspended from the ceiling by a 0.75 m long string.
 

  1. Use labelled arrows to indicate on Figure 8 the two physical forces acting on the ball.   (2 marks)
  1. Calculate the speed of the ball. Show your working.   (4 marks)
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a.  
       

b.    \(1.2\ \text{ms}^{-1}\)

Show Worked Solution

a.    Only two forces are acting on the ball: \(F_w\) and \(F_T\):
 

   

Mean mark 58%.
COMMENT: Many students incorrectly identified centripetal force (this is a result of the tension force).

b.   
     

\(\text{Using}\ \ \tan\theta=\dfrac{F_{\text{net}}}{mg}:\)

\( F_{\text{net}}\) \(=mg\,\tan\theta\)  
\(\dfrac{mv^2}{r}\) \(=mg\,\tan\theta\)  
\(\dfrac{v^2}{r}\) \(=g\,\tan\theta\)  
\(\therefore v\) \(=\sqrt{g\,r\,\tan\theta}\)  
  \(=\sqrt{9.8 \times 0.317 \times \tan 25}\ \ ,\ \ (r = 0.75 \times \sin25=0.317\ \text{m})\)  
  \(=1.2\ \text{ms}^{-1}\)  
♦ Mean mark (b) 50%.

PHYSICS, M6 2020 VCE 6

Two Physics students hold a coil of wire in a constant uniform magnetic field, as shown in Figure 5a. The ends of the wire are connected to a sensitive ammeter. The students then change the shape of the coil by pulling each side of the coil in the horizontal direction, as shown in Figure 5b. They notice a current register on the ammeter.
 

  1. Will the magnetic flux through the coil increase, decrease or stay the same as the students change the shape of the coil?  (1 mark)
  1. Explain, using physics principles, why the ammeter registered a current in the coil and determine the direction of the induced current.  (3 marks
  1. The students then push each side of the coil together, as shown in Figure 6a, so that the coil returns to its original circular shape, as shown in Figure 6b, and then changes to the shape shown in Figure 6c. 
     

  1. Describe the direction of any induced currents in the coil during these changes. Give your reasoning.   (2 marks)

Show Answers Only

a.    Decrease 

b.    Ammeter registers a current:

→ By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current. 

→ This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.

→ Hence by using the right-hand grip rule, the current will run clockwise through the coil.

c.    → There will be an induced current from 6a to 6b.

→ As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.

→ By the right-hand grip rule, the current will flow anti-clockwise around the coil.

→ There will also be an induced current from 6b to 6c.

→ The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.

→ Using the right-hand grip rule, the current through the coil will flow clockwise.

Show Worked Solution

a.    Magnetic flux will decrease.

→ The magnetic flux is proportional to how many field lines are passing through a given area.

→ As the number of lines passing through the coil of wire decreases, the magnetic flux will also decrease.
 

b.    Ammeter registers a current:

→ By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current. 

→ This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.

→ Hence by using the right-hand grip rule, the current will run clockwise through the coil.
 

♦ Mean mark (b) 42%.

c.    → There will be an induced current from 6a to 6b.

→ As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.

→ By the right-hand grip rule, the current will flow anti-clockwise around the coil.

→ There will also be an induced current from 6b to 6c.

→ The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.

→ Using the right-hand grip rule, the current through the coil will flow clockwise.

♦♦♦ Mean mark (c) 28%.
COMMENT: Lenz’s law was poorly understood in this question.

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

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

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

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

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

PHYSICS, M5 2020 VCE 4*

The Ionospheric Connection Explorer (ICON) space weather satellite, constructed to study Earth's ionosphere, was launched in October 2019. ICON will study the link between space weather and Earth's weather at its orbital altitude of 600 km above Earth's surface. Assume that ICON's orbit is a circular orbit. 

  1. Calculate the orbital radius of the ICON satellite.   (1 mark)

  2. Calculate the orbital period of the ICON satellite correct to three significant figures. Show your working.   (4 marks)

  3. Explain how the ICON satellite maintains a stable circular orbit without the use of propulsion engines.   (2 marks)

Show Answers Only

a.    \(r_{orbit}= 6.971 \times 10^6\ \text{m}\)

b.    \(5780\ \text{s}\)

c.    → The icon satellite is only subject to the gravitational force of the Earth.

→ This force is a centripetal force of constant magnitude and hence the satellite maintains a stable circular orbit. 

Show Worked Solution

a.    Radius of orbit is equal to the altitude plus the radius of the Earth.

\(r_{orbit}=600\ 000\ \text{m} + 6.371 \times 10^6\ \text{m} = 6.971 \times 10^6\ \text{m}\)

 

b.     \(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)
  \(T\) \(=\sqrt{\dfrac{4\pi^2r^3}{GM}}\)
    \(=\sqrt{\dfrac{4\pi^2 \times (6.971 \times 10^6)^3}{6.67 \times 10^{-11} \times 6 \times 10^{24}}}\)
    \(=5780\ \text{s}\ \ \text{(3 sig.fig.)}\)

 

c.    → The icon satellite is only subject to the gravitational force of the Earth.

→ This force is a centripetal force of constant magnitude and hence the satellite maintains a stable circular orbit. 

♦♦♦ Mean mark (c) 23%.

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, M6 2020 VCE 3

Electron microscopes use a high-precision electron velocity selector consisting of an electric field, \(E\), perpendicular to a magnetic field, \(B\).

Electrons travelling at the required velocity, \(v_0\), exit the aperture at point \(\text{Y}\), while electrons travelling slower or faster than the required velocity, \(v_0\), hit the aperture plate, as shown in Figure 2.
 

  1. Show that the velocity of an electron that travels straight through the aperture to point \(\text{Y}\) is given by  \( v_{0} \) = \( \dfrac{E}{B}\).   (1 mark)

  2. Calculate the magnitude of the velocity, \(v_0\), of an electron that travels straight through the aperture to point \(\text{Y}\) if  \(E\) = 500 kV m\(^{-1}\)  and  \(B\) = 0.25 T. Show your working.   (2 marks)

  3.  i. At which of the points – \(\text{X, Y}\), or \(\text{Z}\) – in Figure 2 could electrons travelling faster than \(v_0\) arrive?   (1 mark)

  4. ii. Explain your answer to part c.i.   (2 marks)

Show Answers Only

a.    Find \(v_0\) where forces due to magnetic and electric field are balanced

\(F_E\) \(=F_B\)  
\(qE\) \(=qv_0B\)  
\(v_0\) \(=\dfrac{E}{B}\)  

 
b.    \(v_0=2 \times 10^6\ \text{ms}^{-1}\)

c.i.    Point \(\text{Z.}\)

c.ii.  When electrons travel faster than \(v_0\):

→ The force due to the magnetic field will increase but the force due to the electric field will remain unchanged. 

→ Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.

→ Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)

Show Worked Solution

a.    Find \(v_0\) where forces due to magnetic and electric field are balanced

\(F_E\) \(=F_B\)  
\(qE\) \(=qv_0B\)  
\(v_0\) \(=\dfrac{E}{B}\)  
♦ Mean mark (a) 46%.

b.    \(v_0=\dfrac{E}{B}=\dfrac{500\ 000}{0.25}=2 \times 10^6\ \text{ms}^{-1}\)
 

c.i.    Point \(\text{Z.}\)
 

c.ii.  When electrons travel faster than \(v_0\):

→ The force due to the magnetic field will increase but the force due to the electric field will remain unchanged. 

→ Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.

→ Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point \(\text{Z.}\)

♦ Mean mark (c.i.) 40%.
♦♦♦ Mean mark (c.ii.) 15%.
COMMENT: Students needed to include what would happen to the electric force.

PHYSICS, M8 2020 VCE 18 MC

Quantised energy levels within atoms can best be explained by

  1. electrons behaving as individual particles with different energies.
  2. electrons behaving as waves, with each energy level representing a diffraction pattern.
  3. protons behaving as waves, with only standing waves at particular wavelengths allowed.
  4. electrons behaving as waves, with only standing waves at particular wavelengths allowed.
Show Answers Only

\(D\)

Show Worked Solution

→ This was explained by de Broglie’s work in determining electrons (all matter) can exhibit wave properties.

→ Quantised energy levels in the atom correspond to where electrons form standing waves due to there being an integer number of electron wavelengths.

\(\Rightarrow D\)

PHYSICS, M7 2020 VCE 15 MC

\(\text{I}\) The energy of a light wave increases with increasing amplitude.
\(\text{II}\) The energy of a light wave increases with increasing frequency.
\(\text{III}\) The energy of a light wave increases with decreasing wavelength.
Which of the statements above about the energy of light waves is correct?
 
  1. \(\text{III}\) only
  2. \(\text{I}\) and \(\text{II}\) only
  3. \(\text{I}\) and \(\text{III}\) only
  4. all of the statements are correct
Show Answers Only

\(D\)

Show Worked Solution

→ Statement \(\text{I}\) is correct when considering the wave-model of light. The higher the amplitude, the further away the displacement of the wave from the origin which corresponds to higher energy.

→ Statements \(\text{II}\) and \(\text{III}\) are correct when considering the particle model of light.

→ The energy of a light photon, \(E=hf=\dfrac{hc}{\lambda}\). It follows that increasing the frequency or decreasing the wavelength both lead to an increase in energy.

\(\Rightarrow D\)

PHYSICS, M8 2020 VCE 13 MC

Matter is converted to energy by nuclear fusion in stars.

If the star Alpha Centauri converts mass to energy at the rate of 6.6 × 10\(^9\) kg s\(^{-1}\), then the power generated is closest to

  1. \(2.0 \times 10^{18}\ \text{W}\)
  2. \(2.0 \times 10^{18}\ \text{J}\)
  3. \(6.0 \times 10^{26}\ \text{W}\)
  4. \(6.0 \times 10^{26}\ \text{J}\)
Show Answers Only

\(C\)

Show Worked Solution

→ The energy produced by the star each second is the power generated by the star.

→ \(E=mc^2=6.6 \times 10^9 \times (3 \times 10^8)^2=6 \times 10^{26}\ \text{Js}^{-1}=6 \times 10^{26}\ \text{W}\)

\(\Rightarrow C\)

Mean mark 56%.

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%.

Vectors, SPEC2 2022 VCAA 4

A student is playing minigolf on a day when there is a very strong wind, which affects the path of the ball. The student hits the ball so that at time  \(t=0\) seconds it passes through a fixed origin \(O\). The student aims to hit the ball into a hole that is 7 m from \(O\). When the ball passes through \(O\), its path makes an angle of \(\theta\) degrees to the forward direction, as shown in the diagram below.
 

The path of the ball \(t\) seconds after passing through \(O\) is given by

\(\underset{\sim}{\text{r}}(t)=\dfrac{1}{2} \sin \left(\dfrac{\pi t}{4}\right) \underset{\sim}{\text{i}}+2 t \underset{\sim}{\text{j}}\)  for  \(t \in[0,5]\)

where \(\underset{\sim}{i}\) is a unit vector to the right, perpendicular to the forward direction, \(\underset{\sim}{j}\) is a unit vector in the forward direction and displacement components are measured in metres.

  1. Find \(\theta\) correct to one decimal place.   (2 marks)

  2.  i. Find the speed of the ball as it passes through \(O\). Give your answer in metres per second, correct to two decimal places.   (2 marks)

  3. ii. Find the minimum speed of the ball, in metres per second, and the time, in seconds, at which this minimum speed occurs.   (2 marks)

  4. Find the minimum distance from the ball to the hole. Give your answer in metres, correct to three decimal places.   (3 marks)

  5. How far does the ball travel during the first four seconds after passing through \(O\) ? Give your answer in metres, correct to three decimal places.   (2 marks)

Show Answers Only

a.   \(\theta =11.1^{\circ}\)

b.i.  \(2.04\ \text{ms}^{-1}\)

b.ii.  \(\text{Minimum speed} =2 \text{ ms} ^{-1}\)

c.   \(\text {Minimum }\abs{d}=0.188\ \text{m}\)

d.   \(8.077\ \text{m}\)

Show Worked Solution

a.    \(r(t)=\dfrac{1}{2}\,\sin \left(\dfrac{\pi t}{4}\right) \underset{\sim}{i}+2 t \underset{\sim}{j}\)

\(\dot{r}(t)=\dfrac{\pi}{8}\, \cos \left(\dfrac{\pi t}{4}\right) \underset{\sim}{i}+2 \underset{\sim}{j}\)

\(\dot{r}(0)=\dfrac{\pi}{8}\,\underset{\sim}{i}+2\underset{\sim}{j}\)

\begin{aligned}
\tan (90-\theta) & =\dfrac{2}{\frac{\pi}{8}} \\
90-\theta & =\tan ^{-1}\left(\dfrac{16}{\pi}\right)=78.9^{\circ} \\
\theta & =11.1^{\circ}
\end{aligned}

♦ Mean mark (a) 46%.
b.i.     \(\text{Speed}\) \(=\abs{\dot{r}(0)}\)
    \(=\sqrt{\left(\dfrac{\pi}{8}\right)^2+2^2}\)
    \(=2.04\ \text{ms}^{-1}\ \text{(2 d.p.)}\)

 

b.ii.  \(\abs{\dot{r}}=\sqrt{\left(\dfrac{\pi}{8}\right)^2 \times \cos ^2\left(\dfrac{\pi t}{4}\right)+4}\)

 \(\abs{\dot{r}}\text { is a minimum when}\ \ \cos ^2\left(\dfrac{\pi t}{4}\right)=0\)

\(\Rightarrow t=2\)

\(\therefore\ \text{Minimum speed} =\sqrt{4}=2 \text{ ms} ^{-1}\)
 

c.    \(\text{Ball position}\ \Rightarrow \ \underset{\sim}{r}=\dfrac{1}{2}\,\sin \left(\dfrac{\pi t}{4}\right)\underset{\sim}{i}+2 \, t\underset{\sim}{j}\)

\(\text{Hole position}\ \Rightarrow \ \underset{\sim}{h}=0 \underset{\sim}{i}+7\underset{\sim}{j}\)

\(\underset{\sim}{d}=\underset{\sim}{r}-\underset{\sim}{h}=\dfrac{1}{2}\, \sin \left(\dfrac{\pi t}{4}\right) \underset{\sim}{i}+(2 t-7)\underset{\sim}{j}\)

\(\abs{\underset{\sim}{d}}=\sqrt{\left(\dfrac{1}{2}\right)^2 \sin ^2\left(\dfrac{\pi t}{4}\right)+(2 t-7)^2}\)

\(\text {Minimum }\abs{d}=0.188\text{ m (3 d.p.)}\)

♦ Mean mark (c) 45%.
d.     \(\text{Distance}\) \(=\displaystyle \int_0^4\left(\frac{\pi}{8}\, \cos \left(\frac{\pi t}{4}\right)\right)^2+4\, dt\)
    \(=8.077\ \text{m  (3 d.p.)}\)
♦ Mean mark (d) 48%.

CHEMISTRY, M3 EQ-Bank 16

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between lithium hydroxide and hydrochloric acid.   (1 mark)

  2. The reaction between potassium carbonate and hydrochloric acid.   (1 mark)

Show Answers Only

a.    \(\ce{LiOH + HCl -> LiCl + H2O}\)

b.    \(\ce{K2CO3 + 2HCl -> 2KCl + CO2 + H2O}\)

Show Worked Solution

a.    Acid-Base

\[\ce{LiOH + HCl -> LiCl + H2O}\]

b.    Acid-carbonate

\[\ce{K2CO3 + 2HCl -> 2KCl + CO2 + H2O}\]

CHEMISTRY, M3 EQ-Bank 15

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between zinc and sulfuric acid.   (1 mark)

  2. The decomposition of calcium carbonate.   (1 mark)

  3. The incomplete combustion of ethane \(\ce{(C2H6)}\) with a limited amount of oxygen.   (1 mark)

Show Answers Only

a.    \(\ce{Zn + H2SO4 -> ZnSO4 + H2}\)

b.    \(\ce{CaCO3 -> CaO + CO2}\)

c.    \(\ce{2C2H6 + 5O2 -> 4CO + 6H2O}\)

Show Worked Solution

a.    Active metal and acid

\[\ce{Zn + H2SO4 -> ZnSO4 + H2}\]

b.    Decomposition

\[\ce{CaCO3 -> CaO + CO2}\]

c.    Combustion

\[\ce{2C2H6 + 5O2 -> 4CO + 6H2O}\]

CHEMISTRY, M3 EQ-Bank 14

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between potassium hydroxide and sulfuric acid.   (1 mark)
  2. The reaction between sodium bicarbonate \(\ce{(NaHCO3)}\) and acetic acid \(\ce{(CH3COOH)}\).   (1 mark)
Show Answers Only

a.    \(\ce{2KOH + H2SO4 -> K2SO4 + 2H2O}\)

b.    \(\ce{NaHCO3 + CH3COOH -> CH3COONa + CO2 + H2O}\)

Show Worked Solution

a.   Acid-base 

\(\ce{2KOH + H2SO4 -> K2SO4 + 2H2O}\)

b.   Acid-carbonate 

\(\ce{NaHCO3 + CH3COOH -> CH3COONa + CO2 + H2O}\)

CHEMISTRY, M3 EQ-Bank 13

Write balanced chemical equations for each of the following reactions (states of matter are not required).

  1. The reaction between magnesium and hydrochloric acid.   (1 mark)

  2. The decomposition of ammonia.   (1 mark)

  3. The incomplete combustion of butane (\(\ce{C4H10}\)) with a limited amount of oxygen.   (1 mark)

Show Answers Only

a.    \(\ce{Mg + 2HCl -> MgCl2 + H2}\)

b.    \(\ce{2NH3 -> N2 + 3H2}\)

c.    \(\ce{2C4H10 + 9O2 -> 8CO + 10H2O}\)

Show Worked Solution

a.   Reaction between active metal and acid

\(\ce{Mg + 2HCl -> MgCl2 + H2}\)

b.   Decomposition

\(\ce{2NH3 -> N2 + 3H2}\)

c.   Combustion

\(\ce{2C4H10 + 9O2 -> 8CO + 10H2O}\)

CHEMISTRY, M3 EQ-Bank 12

A student stirs 2.80 g of silver \(\text{(I)}\) nitrate powder into 250.0 mL of 1.00 mol L\(^{-1}\) sodium hydroxide solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)

  2. Calculate the theoretical mass of precipitate that will be formed.   (3 marks)

The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).

  1. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)

Show Answers Only

a.    \(\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}\)

b.    \(2.06 \text{ g}\)

c.    Possible reasons for the higher mass:

→ Presence of excess solution in the filter paper or incomplete drying.

→ Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.
 

Experimental adjustment to eliminate error:

→ Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.

→ Dry precipitate completely in the incubator before weighing.

Show Worked Solution

a.    \(\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}\)
 

b.   \(\ce{MM(AgNO3) = 107.9 + 14.01 + 16.00 \times 3 = 169.91}\)

\(\ce{n(AgNO3) = \dfrac{\text{m}}{\text{MM}} = \dfrac{2.80}{169.91} = 0.01648 \text{ mol}}\)

\(\ce{n(NaOH) = c \times V = 1.00 \times 0.250 = 0.250 \text{ mol}}\)

\(\Rightarrow \ce{AgNO3}\ \text{is the limiting reagent}\)

\(\ce{n(AgOH) = n(AgNO3) = 0.01648 \text{ mol}}\)

\(\ce{m(AgOH) = n \times MM = 0.01648 \times (107.9 + 16.00 + 1.008) = 2.06 \text{ g}}\)
 

c.    Possible reasons for the higher mass:

→ Presence of excess solution in the filter paper or incomplete drying.

→ Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.
 

Experimental adjustment to eliminate error:

→ Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.

→ Dry precipitate completely in the incubator before weighing.

CHEMISTRY, M3 EQ-Bank 11

A student stirs 2.45 g of copper \(\text{(II)}\) nitrate powder into 200.0 mL of 1.25 mol L\(^{-1}\) sodium carbonate solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)

  2. Calculate the theoretical mass of precipitate that will be formed.   (3 marks)

The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).

  1. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)

Show Answers Only

a.    \(\ce{Cu(NO3)2(aq) + Na2CO3(aq) -> CuCO3(s) + 2NaNO3(aq)}\)

b.    \(1.61 \text{ g}\)

c.    Possible reasons for the higher mass:

→ Presence of excess solution in the filter paper or incomplete drying.

→ Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.
 

Experimental adjustment to eliminate error:

→ Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.

→ Dry precipitate completely in the incubator before weighing.

Show Worked Solution

a.    \(\ce{Cu(NO3)2(aq) + Na2CO3(aq) -> CuCO3(s) + 2NaNO3(aq)}\)
 

b.   \(\ce{MM(Cu(NO3)2) = 63.55 + 2 \times (14.01 + 16.00 \times 3) = 187.57}\)

\(\ce{n(Cu(NO3)2) = \dfrac{\text{m}}{\text{MM}} = \dfrac{2.45}{187.57} = 0.01306 \text{ mol}}\)

\(\ce{n(Na2CO3) = c \times V = 1.25 \times 0.200 = 0.250 \text{ mol}}\)

\(\Rightarrow \ce{Cu(NO3)2} \text{ is the limiting reagent}\)

\(\ce{n(CuCO3) = n(Cu(NO3)2) = 0.01306 \text{ mol}}\)

\(\ce{m(CuCO3) = n \times MM = 0.01306 \times (63.55 + 12.01 + 16.00 \times 3) = 1.61 \text{ g}}\)
 

c.    Possible reasons for the higher mass:

→ Presence of excess solution in the filter paper or incomplete drying.

→ Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.
 

Experimental adjustment to eliminate error:

→ Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.

→ Dry precipitate completely in the incubator before weighing.

CHEMISTRY, M2 EQ-Bank 8

  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 = \(\ce{CH2O}\)

→ Butyraldehyde and lactic acid 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 = \(\ce{CH2O}\)

→ Butyraldehyde and lactic acid 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 6

During a laboratory experiment, 32.0 grams of oxygen gas \(\ce{(O2)}\) is required. Calculate the number of molecules of oxygen gas needed.   (2 marks)

Show Answers Only

\(6.022 \times 10^{23}\) molecules.

Show Worked Solution

Calculate the number of moles of oxygen gas (\(\ce{O2}\)):

  \(\ce{n(O2) = \dfrac{\text{m}}{\text{MM}} = \dfrac{32.0\ \text{g}}{32.00\ \text{g mol}^{-1}} = 1.00\ \text{mol}}\)
 

Use Avogadro’s number to find the number of molecules:

  \(\ce{N(O2) = n \times N_A = 1.00 \, mol \times 6.022 \times 10^{23} \, mol^{-1} = 6.022 \times 10^{23} \, molecules}\)

→ Oxygen gas required = \(6.022 \times 10^{23}\) molecules.

CHEMISTRY, M2 EQ-Bank 5

Ammonia \(\ce{(NH3)}\) is synthesized from nitrogen \(\ce{(N2)}\) and hydrogen \(\ce{(H2)}\) according to the following equation:

\(\ce{N2(g) + 3H2(g) -> 2NH3(g)}\)

If 1.50 moles of nitrogen gas react completely with hydrogen gas, calculate the mass of ammonia produced.   (2 marks)

Show Answers Only

\(\ce{m(NH3) = 51.0\ \text{g}}\)

Show Worked Solution

Calculate the number of moles of ammonia \(\ce{(NH3)}\) produced:

  \(\ce{n(NH3) = 2 \times n(N2) = 2 \times 1.50 \, mol = 3.00 \, mol}\)
 

Calculate the mass of ammonia produced:

  \(\ce{m(NH3) = n \times MM = 3.00 \, mol \times 17.03\ \text{g mol}^{-1} = 51.09\ \text{g}}\)

→ The mass of ammonia produced is 51.0 grams.

Calculus, MET2 2023 VCE SM-Bank 1

The function \(g\) is defined as follows.

\(g:(0,7] \rightarrow R, g(x)=3\, \log _e(x)-x\)

  1. Sketch the graph of \(g\) on the axes below. Label the vertical asymptote with its equation, and label any axial intercepts, stationary points and endpoints in coordinate form, correct to three decimal places.   (3 marks)

     

     

  2.  i. Find the equation of the tangent to the graph of \(g\) at the point where \(x=1\).   (1 mark)

  3. ii. Sketch the graph of the tangent to the graph of \(g\) at \(x=1\) on the axes in part a.   (1 mark)

Newton's method is used to find an approximate \(x\)-intercept of \(g\), with an initial estimate of \(x_0=1\).

  1. Find the value of \(x_1\).   (1 mark)

  2. Find the horizontal distance between \(x_3\) and the closest \(x\)-intercept of \(g\), correct to four decimal places.   (1 mark)

  3.  i. Find the value of \(k\), where \(k>1\), such that an initial estimate of  \(x_0=k\)  gives the same value of  \(x_1\)  as found in part \(c\). Give your answer correct to three decimal places.   (2 marks)

  4. ii. Using this value of \(k\), sketch the tangent to the graph of \(g\) at the point where  \(x=k\)  on the axes in part a.   (1 mark)

Show Answers Only
a.    
  
b.i

 \(y=2x-3\)
  
b.ii
  
c. \(\dfrac{3}{2}=1.5\)
  
d. \(0.0036\)
  
e.i \(k=2.397\)
  
e.ii
Show Worked Solution

a.

b.i    \(g(x)\) \(=3\log_{e}x-x\)
  \(g(1)\) \(=3\log_{e}1-1=-1\)
  \(g^{\prime}(x)\) \(=\dfrac{3}{x}-1\)
  \(g^{\prime}(1)\) \(=\dfrac{3}{1}-1=2\)

  
\(\text{Equation of tangent at }(1, -1)\ \text{with }m=2\)

\(y+1=2(x-1)\ \ \rightarrow \ \ y=2x-3\)

b.ii

 
c.    \(\text{Newton’s Method}\)

\(x_1\) \(=x_0-\dfrac{g(x)}{g'(x)}\)
  \(=1-\left(\dfrac{-1}{2}\right)\)
  \(=\dfrac{3}{2}=1.5\)

\(\text{Using CAS:}\)
  
 

d.    \(\text{Using CAS}\)

\(x\text{-intercept}:\ x=1.85718\)

\(\therefore\ \text{Horizontal distance}=1.85718-1.85354=0.0036\)

 

e.i.  \(\text{Using CAS}\)

\(k-\dfrac{3\log_{e}x-x}{\dfrac{3}{x}-1}\) \(=1.5\)
\(k>1\ \therefore\ \ k\) \(=2.397\)

 
e.ii

CHEMISTRY, M2 EQ-Bank 2

During a laboratory experiment, a gas is collected in a sealed syringe. Initially, the gas has a volume of 5.0 litres and a pressure of 1.0 atmosphere. 

  1.  Calculate the new pressure inside the syringe when the volume is decreased to 3.0 litres, assuming no temperature change.   (2 marks)

  2. After reaching the pressure calculated in part a, the volume is further decreased so that the pressure inside the syringe doubles. Calculate the final volume of the gas.   (2 marks)

  3. Discuss two potential experimental errors that could affect the accuracy of the observed results compared to the theoretical predictions of Boyle's Law.   (2 marks)

Show Answers Only

a.    \(1.67\ \text{atm}\)

b.    \(1.5\ \text{L}\)

c.    Potential experimental errors could include:

→ Temperature Control. Any temperature changes violate the assumptions of Boyle’s Law which holds only at constant temperature.

→ Measurement Accuracy. Errors in measuring volume changes can lead to inaccuracies in pressure calculations.

→ Non-Ideal Behaviour. At high pressures, gases may deviate from ideal behaviour, affecting the accuracy of Boyle’s Law predictions.

Show Worked Solution

a.    Using Boyle’s Law  \( P_1V_1 = P_2V_2 \):

\( P_1 = 1.0 \, \text{atm}, \quad V_1 = 5.0 \, \text{L}, \quad V_2 = 3.0 \, \text{L} \)

\(P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{1.0 \times 5.0}{3.0} = 1.67 \, \text{atm} \)
 

b.    To find the new volume when pressure doubles:

\( P_3 = 2 \times P_2 = 2 \times 1.67 \, \text{atm} = 3.34 \, \text{atm}\)

\( P_1V_1 = P_3V_3 \ \  \Rightarrow \ \  V_3 = \dfrac{P_1 \times V_1}{P_3} = \dfrac{1.0  \times 5.0 }{3.34} \approx 1.5 \, \text{L}\)
 

c.    Potential experimental errors could include:

→ Temperature Control. Any temperature changes violate the assumptions of Boyle’s Law which holds only at constant temperature.

→ Measurement Accuracy. Errors in measuring volume changes can lead to inaccuracies in pressure calculations.

→ Non-Ideal Behaviour. At high pressures, gases may deviate from ideal behaviour, affecting the accuracy of Boyle’s Law predictions.

CHEMISTRY, M2 EQ-Bank 1

A cylinder equipped with a movable piston contains an inert gas. The initial pressure of the gas is 1.5 atmospheres, and the cylinder's initial volume is 4.0 litres.

The piston is then adjusted to change the volume of the cylinder.

Assuming the temperature of the gas remains constant throughout the process, calculate the new pressure of the gas when the volume is adjusted to 2.0 litres.   (2 marks)

Show Answers Only

\(3.0\ \text{atm}\)

Show Worked Solution

\(P_1 = 1.5\ \text{atm},\ \ V_1 = 4.0\ \text{L},\ \ V_2 = 2.0\ \text{L} \)

Boyle’s Law  \(\Rightarrow \ P_1V_1 = P_2V_2 \).

\( P_2 = \dfrac{P_1 \times V_1}{V_2} = \dfrac{1.5 \, \text{atm} \times 4.0 \, \text{L}}{2.0 \, \text{L}} = 3.0 \, \text{atm}\)

CHEMISTRY, M2 EQ-Bank 4

A cylinder equipped with a movable piston contains an inert gas. The initial pressure of the gas is 1.35 atmospheres, and the cylinder's initial volume is 5.0 litres.

The piston is then adjusted to change the volume of the cylinder.

Assuming the temperature of the gas remains constant throughout the process, calculate the new pressure of the gas when the volume of the cylinder is adjusted to 8.0 litres.   (2 marks)

Show Answers Only

\(0.84\ \text{atm}\)

Show Worked Solution

\(V_1 = 5.0\ \text{L},\ \ V_2 = 8.0\ \text{L},\ \ P_1=1.35\ \text{atm} \)

Boyle’s Law  \(\Rightarrow \ P_1V_1 = P_2V_2 \).

\(P_2 = \dfrac{P_1 \times V_1}{V_{2}} = \dfrac{1.35\ \text{atm} \times 5.0\ \text{L}}{8.0\ \text{L}} = 0.84\ \text{atm}\)

CHEMISTRY, M1 EQ-Bank 36

A mixture of sand and salt was provided to a group of students for them to determine its percentage composition by mass.

They added water to the sample before using filtration and evaporation to separate the components.

During the evaporation step, the students noticed white powder ‘spitting’ out of the basin onto the bench, so they turned off the Bunsen burner and allowed the remaining water to evaporate overnight.

After filtering, they allowed the filter paper to dry overnight before weighing. An electronic balance was used to measure the mass of each component to two decimal places.

The results were recorded as shown:

    • Mass of the original sand and salt mixture = 15.73 g
    • Mass of the filter paper = 0.80 g
    • Mass of the dried filter paper after filtering = 11.95 g
    • Mass of the empty evaporating basin = 33.50 g
    • Mass of the evaporating basin after evaporation = 36.60 g
  1. Calculate the percentage composition by mass of sand AND salt in the mixture.   (3 marks)

  2. Consider the following definition of validity
  3. Validity is the degree to which tests measure what was intended, or the accuracy of actions, data and inferences produced from tests and other processes.
  4. Use this definition of to assess the validity of the experiment.   (2 marks)

Show Answers Only

a.    % Sand = 70.88%

% Salt = 19.71%

b.   → The calculations show that the percentages do not add up to 100.

→ Some mass (salt) was observed to be lost during the experiment, and thus the mass of salt determined is lower than the true value and the results are not accurate.

→ Hence, experiment is not valid.

Show Worked Solution
a.     \(\text{Sand mass}\ \) \(\text{ = Mass of dried filter paper – Mass of filter paper}\)
    \(= 11.95-0.80 = 11.15\ \text{g}\)
\(\text{Salt mass}\) \(\text{ = Mass of dried filter paper – Mass of filter paper}\)  
  \(= 36.60-33.50 = 3.10\ \text{g}\)  

 

\(\text{% sand} = \left(\dfrac{\text{Mass of sand}}{\text{Mass of original mixture}}\right) \times 100= \left(\dfrac{11.15 \ \text{g}}{15.73 \ \text{g}}\right) \times 100 = 70.88\%\)

\(\text{% salt} = \left(\dfrac{\text{Mass of salt}}{\text{Mass of original mixture}}\right) \times 100 = \left(\dfrac{3.10 \ \text{g}}{15.73 \ \text{g}}\right) \times 100 = 19.71\%\)
 

b.   → The calculations show that the percentages do not add up to 100.

→ Some mass (salt) was observed to be lost during the experiment, and thus the mass of salt determined is lower than the true value and the results are not accurate.

→ Hence, experiment is not valid.

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 32

Discuss the reasons why Neon, similar to other noble gases, does not easily form chemical bonds with other elements.

Explain Neon's electron configuration and how this relates to its position on the periodic table and its chemical characteristics.   (3 marks)

Show Answers Only

→ Neon is located in Group 18 of the periodic table and has an atomic number of 10.

→ Its electron configuration is 1s² 2s² 2p⁶. This configuration shows that Neon has a completely filled outer shell, a characteristic of noble gases, which contributes to its chemical inertness.

→ As a member of Group 18, Neon shares this fully filled valence shell characteristic with other noble gases, resulting in high stability and very low reactivity.

Show Worked Solution

→ Neon is located in Group 18 of the periodic table and has an atomic number of 10.

→ Its electron configuration is 1s² 2s² 2p⁶. This configuration shows that Neon has a completely filled outer shell, a characteristic of noble gases, which contributes to its chemical inertness.

→ As a member of Group 18, Neon shares this fully filled valence shell characteristic with other noble gases, resulting in high stability and very low reactivity.

CHEMISTRY, M1 EQ-Bank 29

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. Discuss 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.

Probability, MET2 2023 VCE SM-Bank 4 MC

The probability density function \(f\) of a random variable \(X\) is given by

\(f(x)= \begin{cases}\dfrac{x+1}{20} & 0 \leq x<4 \\ \dfrac{36-5 x}{64} & 4 \leq x \leq 7.2 \\ 0 & \text {elsewhere}\end{cases}\)

The value of \(a\) such that \(\text{Pr}(X \leq a)=\dfrac{5}{8}\) is

  1. \(\dfrac{4(\sqrt{15}-9)}{5}\)
  2. \(\sqrt{26}-1\)
  3. \(\dfrac{36-4 \sqrt{15}}{5}\)
  4. \(\dfrac{4 \sqrt{15}+9}{5}\)
  5. \(\dfrac{4 \sqrt{15}+36}{5}\)
Show Answers Only

 

Show Worked Solution

\(\text{Using CAS:}\)

\(\displaystyle \int_{0}^{4}\dfrac{x+1}{20}\,dx =\dfrac{3}{5}<\dfrac{5}{8}\)

\(\therefore\ \text{Pr}(0\leq X<4)+\text{Pr}(4\leq X<a)=\dfrac{5}{8}\)

\(\rightarrow\quad\) \(\dfrac{3}{5}+\displaystyle \int_{4}^{a}\dfrac{36-5x}{64}\,dx\)  \(=\dfrac{5}{8}\)
  \(\displaystyle \int_{4}^{a}\dfrac{36-5x}{64}\,dx\) \(=\dfrac{5}{8}-\dfrac{3}{5}=\dfrac{1}{40}\)
  \(\therefore\ \text{from CAS:}\quad \ a\)

\(=\dfrac{-4\Big(\sqrt{15}-9\Big)}{5}\)
   

\(=\dfrac{36-4\sqrt{15}}{5}\)

  
\(\text{Using CAS:}\)

  
\(\Rightarrow C\)

Calculus, MET2 2023 VCE SM-Bank 3 MC

The area between the curve  \(\displaystyle y=\frac{1}{27}(x-3)^2(x+3)^2+1\)  and the \(x\)-axis on the interval \(x \in[0,4]\) has been approximated using the trapezium rule, as shown in the graph below.

Using the trapezium rule, the approximate area calculated is equal to

  1. \(\displaystyle \frac{1}{2}\left(4+\frac{91}{27}+\frac{52}{27}+1+\frac{76}{27}\right)\)
  2. \(\displaystyle \frac{1}{2}\left(4+\frac{182}{27}+\frac{104}{27}+2+\frac{76}{27}\right)\)
  3. \(\displaystyle \frac{1}{2}\left(8+\frac{182}{27}+\frac{104}{27}+2+\frac{152}{27}\right)\)
  4. \(\displaystyle \frac{1}{2}\left(\frac{182}{27}+\frac{104}{27}+2+\frac{76}{27}\right)\)
  5. \(\displaystyle \frac{1}{2}\left(8+\frac{182}{27}+\frac{104}{27}+2\right)\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Area each trapezium}=\dfrac{h}{2}(f(a)+f(b))\)

\(A\) \(=\Bigg(\dfrac{1}{2}(f(0)+f(1))+\dfrac{1}{2}(f(1)+f(2))+\dfrac{1}{2}(f(2)+f(3))+\dfrac{1}{2}(f(3)+f(4)\Bigg)\)
  \(=\dfrac{1}{2}\Bigg(f(0)+2f(1)+2f(2)+2f(3)+f(4)\Bigg)\)
  \(=\dfrac{1}{2}\Bigg(4+2\times\dfrac{91}{27}+2\times\dfrac{52}{27}+2\times 1+\dfrac{76}{27}\Bigg)\)
  \(=\dfrac{1}{2}\Bigg(4+\dfrac{182}{27}+\dfrac{104}{27}+2+\dfrac{76}{27}\Bigg)\)

  
\(\Rightarrow B\)

Algebra, MET2 2023 VCE SM-Bank 1 MC

\begin{aligned}
& x-2 y=3 \\
& 2 y-z=4
\end{aligned}

Which one of the following correctly describes the general solution to the system of linear equations given above?

  1. \(x=k, \quad y=\dfrac{1}{2}(k+3), \ z=k-1\), for all \(k \in R\)
  2. \(x=k, \quad y=\dfrac{1}{2}(k+3), \ z=k+1\), for all \(k \in R\)
  3. \(x=k, \quad y=\dfrac{1}{2}(k-3), \ z=k+7\), for all \(k \in R\)
  4. \(x=k, \quad y=\dfrac{1}{2}(k-3), \ z=k-7\), for all \(k \in R\)
  5. \(x=k, \quad y=\dfrac{1}{2}(k+3), \ z=k-7\), for all \(k \in R\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{When }x=k\quad (1)\)

\(2y\) \(=k-3\)
\(y\) \(=\dfrac{1}{2}(k-3)\quad (2)\)
\(\text{and}\ z\) \(=2y-4\)
\(z\) \(=2\times\dfrac{1}{2}(k-3)-4\)
  \(=k-7\quad (3)\)

  
\(\text{Using CAS:}\)

\(\Rightarrow D\)

Calculus, MET2 2022 VCAA 5

Consider the composite function `g(x)=f(\sin (2 x))`, where the function `f(x)` is an unknown but differentiable function for all values of `x`.

Use the following table of values for `f` and `f^{\prime}`.

`\quad x \quad` `\quad\quad 1/2\quad\quad` `\quad\quad(sqrt{2})/2\quad\quad` `\quad\quad(sqrt{3})/2\quad\quad`
`f(x)` `-2` `5` `3`
`\quad\quad f^{prime}(x)\quad\quad` `7` `0` `1/9`

 

  1. Find the value of `g\left(\frac{\pi}{6}\right)`.   (1 mark)

The derivative of `g` with respect to `x` is given by `g^{\prime}(x)=2 \cdot \cos (2 x) \cdot f^{\prime}(\sin (2 x))`.

  1. Show that `g^{\prime}\left(\frac{\pi}{6}\right)=\frac{1}{9}`.   (1 mark)

  2. Find the equation of the tangent to `g` at `x=\frac{\pi}{6}`.   (2 marks)

  3. Find the average value of the derivative function `g^{\prime}(x)` between `x=\frac{\pi}{8}` and `x=\frac{\pi}{6}`.   (2 marks)

  4. Find four solutions to the equation `g^{\prime}(x)=0` for the interval `x \in[0, \pi]`.   (3 marks)

Show Answers Only

a.    `3`

b.    `1/9`

c.    `y=1/9x+3-pi/54`

d.    `-48/pi`

e.    ` x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`

Show Worked Solution
 

a.  `g(pi/6)`
 `= f(sin(pi/3))`  
  `= f(sqrt3/2)`  
  `= 3`  

 

b.  `g\ ^{prime}(x)` `= 2\cdot\ cos(pi/3)\cdot\ f\ ^{prime}(sin(pi/3))`  
`g\ ^{prime}(pi/6)` `= 2 xx 1/2 xx f\ ^{prime}(sqrt3/2)`  
  `= 1/9`  

 

c.   `m = 1/9`  and  `g(pi/6) = 3`

`y  –  y_1` `= m(x-x_1)`  
`y  –  3` `= 1/9(x-pi/6)`  
`y` `= 1/9x + 3-pi/54`  

♦♦ Mean mark (c) 45%.
MARKER’S COMMENT: Some students did not produce an equation as required.

  
d.   The average value of `g^{\prime}(x)` between `x=\frac{\pi}{8}` and `x=\frac{\pi}{6}`

Average `= \frac{1}{\frac{\pi}{6}-\frac{\pi}{8}}\cdot\int_{\frac{\pi}{8}}^{\frac{\pi}{6}} g^{\prime}(x) d x`  
  `=24/pi \cdot[g(x)]_{\frac{\pi}{8}}^{\frac{\pi}{\6}}`  
  `= 24/pi \cdot(f(sqrt3/2)-f(sqrt2/2))`  
  `= 24/pi (3-5) = -48/pi`  

♦♦ Mean mark (d) 30%.
MARKER’S COMMENT: Those who used the Average Value formula were generally successful.
Some students substituted `g^{\prime}(x)`, not `g(x)`.

e.   `2 \cos (2 x) f^{\prime}(\sin (2 x)) = 0`

`:.\   2 \cos (2 x) = 0\ ….(1)`  or  ` f^{\prime}(\sin (2 x)) = 0\ ….(2)`

(1):   ` 2 \cos (2 x)`  `= 0`      `x \in[0, \pi]`
`\cos (2 x)` `= 0`      `2 x \in[0,2 \pi]`
`2x` `= pi/2 , (3pi)/2`  
`x` `= pi/4 , (3pi)/4`  
     
(2): ` f^{\prime}(\sin (2 x)) ` `= sqrt2/2`  
`2x` `= pi/4 , (3pi)/4`  
`x` `= pi/8 , (3pi)/8`  

  
`:. \  x = pi/8 , pi/4 , (3pi)/8 ,(3pi)/4`


♦♦ Mean mark (e) 30%.
MARKER’S COMMENT: Some students were able to find `pi/4, (3pi)/4`. Some solved `2 cos(2x)=0` or `f^{\prime}(sin(2x))=0` but not both.

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%.

Probability, MET2 2022 VCAA 3

Mika is flipping a coin. The unbiased coin has a probability of \(\dfrac{1}{2}\) of landing on heads and \(\dfrac{1}{2}\) of landing on tails.

Let \(X\) be the binomial random variable representing the number of times that the coin lands on heads.

Mika flips the coin five times.

    1. Find \(\text{Pr}(X=5)\).   (1 mark)
    2. Find \(\text{Pr}(X \geq 2).\) (1 mark)
    3. Find \(\text{Pr}(X \geq 2 | X<5)\), correct to three decimal places.   (2 marks)
    4. Find the expected value and the standard deviation for \(X\).   (2 marks)

The height reached by each of Mika's coin flips is given by a continuous random variable, \(H\), with the probability density function
  

\(f(h)=\begin{cases} ah^2+bh+c         &\ \ 1.5\leq h\leq 3 \\ \\ 0       &\ \ \text{elsewhere} \\ \end{cases}\)
  

where \(h\) is the vertical height reached by the coin flip, in metres, between the coin and the floor, and \(a, b\) and \(c\) are real constants.

    1. State the value of the definite integral \(\displaystyle\int_{1.5}^3 f(h)\,dh\).   (1 mark)
    2. Given that  \(\text{Pr}(H \leq 2)=0.35\)  and  \(\text{Pr}(H \geq 2.5)=0.25\), find the values of \(a, b\) and \(c\).   (3 marks)

    3. The ceiling of Mika's room is 3 m above the floor. The minimum distance between the coin and the ceiling is a continuous random variable, \(D\), with probability density function \(g\).
    4. The function \(g\) is a transformation of the function \(f\) given by \(g(d)=f(rd+s)\), where \(d\) is the minimum distance between the coin and the ceiling, and \(r\) and \(s\) are real constants.
    5. Find the values of \(r\) and \(s\).   (1 mark)

  1. Mika's sister Bella also has a coin. On each flip, Bella's coin has a probability of \(p\) of landing on heads and \((1-p)\) of landing on tails, where \(p\) is a constant value between 0 and 1 .
  2. Bella flips her coin 25 times in order to estimate \(p\).
  3. Let \(\hat{P}\) be the random variable representing the proportion of times that Bella's coin lands on heads in her sample.
    1. Is the random variable \(\hat{P}\) discrete or continuous? Justify your answer.   (1 mark)
    2. If \(\hat{p}=0.4\), find an approximate 95% confidence interval for \(p\), correct to three decimal places.   (1 mark)
    3. Bella knows that she can decrease the width of a 95% confidence interval by using a larger sample of coin flips.
    4. If \(\hat{p}=0.4\), how many coin flips would be required to halve the width of the confidence interval found in part c.ii.?   (1 mark)

Show Answers Only

a. i.    `frac{1}{32}`    ii.    `frac{13}{16}`    iii.    `0.806` (3 d.p.)

a. iv    `text{E}(X)=5/2,  text{sd}(X)=\frac{\sqrt{5}}{2}`

b. i.   `1`   

b. ii.    `a=-frac{4}{5},  b=frac{17}{5},  c=-frac{167}{60}`

b. iii. `r=-1,  s=3`

c. i.   `text{Discrete}`   ii.   `(0.208,  0.592)`   iii.   `n=100`

Show Worked Solution

a.i  `X ~ text{Bi}(5 , frac{1}{2})`

`text{Pr}(X = 5) = (frac{1}{2})^5 = frac{1}{32}`
 

a.ii  By CAS:    `text{binomCdf}(5,0.5,2,5)`     `0.8125`

`text{Pr}(X>= 2) = 0.8125 = frac{13}{16}`

  
a.iii  `\text{Pr}(X \geq 2 | X<5)`

`=frac{\text{Pr}(2 <= X < 5)}{\text{Pr}(X < 5)} = frac{\text{Pr}(2 <= X <= 4)}{text{Pr}(X <= 4)}`

  
By CAS: `frac{text{binomCdf}(5,0.5,2,4)}{text{binomCdf}(5,0.5,0,4)}`
 

`= 0.806452 ~~ 0.806` (3 decimal places)
  

a.iv `X ~ text{Bi}(5 , frac{1}{2})`

`text{E}(X) = n xx p= 5 xx 1/2 = 5/2`

`text{sd}(X) =\sqrt{n p(1-p)}=\sqrt((5/2)(1 – 0.5)) = \sqrt{5/4} =\frac{\sqrt{5}}{2}`

  

b.i   `\int_{1.5}^3 f(h) d h = 1`


♦♦ Mean mark (b.i) 40%.
MARKER’S COMMENT: Many students did not evaluate the integral or evaluated incorrectly.

b.ii  By CAS:

`f(h):= a\·\h^2 + b\·\h +c`
 

`text{Solve}( {(\int_{1.5}^2 f(h) d h = 0.35), (\int_{2.5}^3 f(h) d h = 0.25), (\int_{1.5}^3 f(h) d h = 1):})`

`a = -0.8 = frac{-4}{5}, \ b = 3.4 = frac{17}{5},\  c = =-2.78 \dot{3} = frac{-167}{60}`


♦♦ Mean mark (b.ii) 47%.
MARKER’S COMMENT: Many students did not give exact answers.

b.iii  `h + d = 3`

`:.\  f(h) = f(3  –  d) = f(- d + 3)`

`:.\  r = – 1 ` and ` s = 3`


♦♦♦♦ Mean mark (b.iii) 10%.
MARKER’S COMMENT: Many students did not attempt this question.

c.i  `\hat{p}`  is discrete.

The number of coin flips must be zero or a positive integer so  `\hat{p}`  is countable and therefore discrete.
 

c.ii  `\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(0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\ , 0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\right)`

`\approx(0.208\ ,0.592)`

 

c.iii To halve the width of the confidence interval, the standard deviation needs to be halved.

`:.\  \frac{1}{2} \sqrt{\frac{0.4 \times 0.6}{25}} = \sqrt{\frac{0.4 \times 0.6}{4 xx 25}} = \sqrt{\frac{0.4 \times 0.6}{100}}`

`:.\  n = 100`

She would need to flip the coin 100 times


♦♦♦ Mean mark (c.iii) 30%.
MARKER’S COMMENT: Common incorrect answers were 0, 10, 11, 50 and 101.

Functions, 2ADV F1 SM-Bank 25

Show that the parabola  \(2x^2-kx+k-2\)  has at least one real root.  (3 marks)

Show Answers Only

\(2x^2-kx+k-2=0\)

\(\Delta=b^2-4ac=(-k)^2-4 \times 2(k-2) = k^{2}-8k+16\)

\(\text{Real roots:}\ \ \Delta \geqslant 0\)

\(k^2-8k+16\) \(\geqslant 0\)  
\((x-4)^2\) \(\geqslant 0\)  

 
\(\therefore\ \text{At least one root exists for all}\ k\)

Show Worked Solution

\(2x^2-kx+k-2=0\)

\(\Delta=b^2-4ac=(-k)^2-4 \times 2(k-2) = k^{2}-8k+16\)

\(\text{Real roots:}\ \ \Delta \geqslant 0\)

\(k^2-8k+16\) \(\geqslant 0\)  
\((x-4)^2\) \(\geqslant 0\)  

 
\(\therefore\ \text{At least one root exists for all}\ k\)

CHEMISTRY, M1 EQ-Bank 27

Silver oxide, when heated, decomposes according to the following chemical equation:

\(\ce{2Ag2O -> 4Ag + O2}\)

231.74 grams of silver oxide (\(\ce{Ag2O}\)) is heated, and the metallic silver (\(\ce{Ag}\)) residue is weighed and recorded in the table below.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag2O} \rule[-1ex]{0pt}{0pt} & \text{231.74 grams} \\
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag} \rule[-1ex]{0pt}{0pt} & \text{216.0 grams} \\
\hline
\end{array}

Calculate the percentage of oxygen (\(\ce{O2}\)), by weight, released from the silver oxide during this process.   (2 marks)

Show Answers Only

\(\%\ce{O2} = 6.8\%\)

Show Worked Solution

\(\text{Initial mass of}\ \ce{Ag2O} = 231.74\ \text{grams}\)

\(\text{Mass of}\ \ce{Ag} = 216.0\ \text{grams}\)

\(\text{Mass of}\ \ce{O2} = 231.74-216.0 = 15.74\ \text{grams}\)

\(\%\ce{O2}\ \text{released}\ = \left(\dfrac{15.74}{231.74} \times 100\%\right) = 6.8\%\)

Complex Numbers, SPEC2 2022 VCAA 2

Two complex numbers \(u\) and \(v\) are given by  \(u=a+i\)  and  \(v=b-\sqrt{2}i\), where \(a, b \in R\).

  1.  i. Given that  \(uv=(\sqrt{2}+\sqrt{6})+(\sqrt{2}-\sqrt{6})i\), show that  \(a^2+(1-\sqrt{3}) a-\sqrt{3}=0\).   (2 marks)

  2. ii. One set of possible values for \(a\) and \(b\) is  \(a=\sqrt{3}\)  and  \(b=\sqrt{2}\).
  3.     Hence, or otherwise, find the other set of possible values.   (1 mark)

  4. Plot and label the points representing  \(u=\sqrt{3}+i\)  and  \(v=\sqrt{2}-\sqrt{2}i\)  on the Argand diagram below.
     

  1. The ray given by  \(\text{Arg}(z)=\theta\)  passes through the midpoint of the line interval that joins the points  \(u=\sqrt{3}+i\)  and  \(v=\sqrt{2}-\sqrt{2}i\).
  2. Find, in radians, the value of \(\theta\) and plot this ray on the Argand diagram in part b.   (2 marks)

  3. The line interval that joins the points  \(u=\sqrt{3}+i\)  and  \(v=\sqrt{2}-\sqrt{2}i\)  cuts the circle  \(|z|=2\)  into a major and a minor segment.
  4. Find the area of the minor segment, giving your answer correct to two decimal places.   (2 marks)

Show Answers Only

a.i.  \(\text{See worked solutions.}\)

a.ii.  \(a=-1, \ b=-\sqrt{6}\)

b.   
       

c.   \(\theta=-\dfrac{\pi}{24}\)

d.    \(0.69\ \text{u}^2\)

Show Worked Solution

a.i.  \(u v=(\sqrt{2}+\sqrt{6})+(\sqrt{2}-\sqrt{6}) i\)

\begin{aligned}
(a+i)(b-\sqrt{2} i) & =a b-\sqrt{2} a i+b i+\sqrt{2} \\
& =a b+\sqrt{2}+(b-\sqrt{2}a) i
\end{aligned}

\(\text {Equating co-efficients: }\)

  \(ab=\sqrt{6} \ \Rightarrow \ b=\dfrac{\sqrt{6}}{a}\ \cdots\ \text {(1)}\)

  \(b-\sqrt{2} a=\sqrt{2}-\sqrt{6}\ \cdots\ \text {(2)}\)

\(\text{Substitute (1) into (2):}\)

\(\dfrac{\sqrt{6}}{a}-\sqrt{2}a=\sqrt{2}-\sqrt{6}\)

\(\sqrt{6}-\sqrt{2} a^2=(\sqrt{2}-\sqrt{6}) a\)

\(\sqrt{2} a^2+(\sqrt{2}-\sqrt{6}) a-\sqrt{6}\) \(=0\)  
\(a^2+(1-\sqrt{3}) a-\sqrt{3}\) \(=0\)  
Mean mark (a) 54%.

 
a.ii. \(\text{Solve } a^2+(1-\sqrt{3}) a-\sqrt{3}=0 \ \ \text{for}\  a :\)

\(a=\sqrt{3} \ \text{ or } -1\)

\(\therefore \text{ Other solution: } a=-1, \ b=-\sqrt{6}\)
 

b.   
       


c.    		 \begin{aligned}
\theta & =\frac{\operatorname{Arg}(u)-\operatorname{Arg}(v)}{2} \\
& =\frac{1}{2}\left(\frac{\pi}{6}-\frac{\pi}{4}\right) \\
& =-\frac{\pi}{24}
\end{aligned}
♦ Mean mark (c) 39%.

d.   \(\text {Sector angle (centre) }=\dfrac{\pi}{6}+\dfrac{\pi}{4}=\dfrac{5 \pi}{12}\)

\(\text {Area of sector }=\dfrac{\frac{5 \pi}{12}}{2 \pi} \times \pi \times 2^2=\dfrac{5 \pi}{6}\)

\begin{aligned}
\text {Area of segment } & =\frac{5 \pi}{6}-\dfrac{1}{2} \times 2 \times 2 \times \sin \left(\dfrac{5 \pi}{6}\right) \\
& \approx 0.69\ \text{u} ^2
\end{aligned}

♦ Mean mark (d) 40%.

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, EXT2 V1 2022 SPEC2 12 MC

Consider the vectors  \(\underset{\sim}{\text{u}}(x)=-\text{cosec}(x) \underset{\sim}{\text{i}}+\sqrt{3} \underset{\sim}{\text{j}}\)  and  \(\underset{\sim}{\text{v}}(x)=\text{cos}(x) \underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}\).

If \(\underset{\sim}{\text{u}}(x)\) is perpendicular to \(\underset{\sim}{\text{v}}(x)\), then possible values for \(x\) are

  1. \(\dfrac{\pi}{6}\) and \(\dfrac{7 \pi}{6}\)
  2. \(\dfrac{\pi}{3}\) and \(\dfrac{4 \pi}{3}\)
  3. \(\dfrac{5 \pi}{6}\) and \(\dfrac{11 \pi}{6}\)
  4. \(\dfrac{2 \pi}{3}\) and \(\dfrac{5 \pi}{3}\)
Show Answers Only

\(A\)

Show Worked Solution

\(\underset{\sim}{u} \perp \underset{\sim}{v} \ \Rightarrow \ \underset{\sim}{u} \cdot \underset{\sim}{v}=0\)

\(-\text{cosec}(x) \cos (x)+\sqrt{3}=0\)

\begin{aligned}
-\dfrac{\cos (x)}{\sin (x)}+\sqrt{3} & =0 \\
\tan (x) & =\dfrac{1}{\sqrt{3}}
\end{aligned}

\(x=\dfrac{\pi}{6}, \dfrac{7 \pi}{6}\)

\(\Rightarrow A\)

Vectors, SPEC2 2022 VCAA 12 MC

Consider the vectors  \(\underset{\sim}{\text{u}}(x)=-\text{cosec}(x) \underset{\sim}{\text{i}}+\sqrt{3} \underset{\sim}{\text{j}}\)  and  \(\underset{\sim}{\text{v}}(x)=\text{cos}(x) \underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}\).

If \(\underset{\sim}{\text{u}}(x)\) is perpendicular to \(\underset{\sim}{\text{v}}(x)\), then possible values for \(x\) are

  1. \(\dfrac{\pi}{6}\) and \(\dfrac{7 \pi}{6}\)
  2. \(\dfrac{\pi}{3}\) and \(\dfrac{4 \pi}{3}\)
  3. \(\dfrac{5 \pi}{6}\) and \(\dfrac{11 \pi}{6}\)
  4. \(\dfrac{2 \pi}{3}\) and \(\dfrac{5 \pi}{3}\)
  5. \(\dfrac{\pi}{6}\) and \(\dfrac{5 \pi}{6}\)
Show Answers Only

\(A\)

Show Worked Solution

\(\underset{\sim}{u} \perp \underset{\sim}{v} \ \Rightarrow \ \underset{\sim}{u} \cdot \underset{\sim}{v}=0\)

\(-\text{cosec}(x) \cos (x)+\sqrt{3}=0\)

\begin{aligned}
-\dfrac{\cos (x)}{\sin (x)}+\sqrt{3} & =0 \\
\tan (x) & =\dfrac{1}{\sqrt{3}}
\end{aligned}

\(\text {Solve by (CAS)}: \ x=\dfrac{\pi}{6}, \dfrac{7 \pi}{6}\)

\(\Rightarrow A\)

Vectors, SPEC2 2022 VCAA 11 MC

Consider the vectors  \(\underset{\sim}{\text{a}}=2 \underset{\sim}{\text{i}}-3 \underset{\sim}{\text{j}}+p \underset{\sim}{\text{k}}, \ \underset{\sim}{\text{b}}=\underset{\sim}{\text{i}}+2 \underset{\sim}{\text{j}}-q \underset{\sim}{\text{k}}\)  and  \(\underset{\sim}{\text{c}}=-3 \underset{\sim}{\text{i}}+2 \underset{\sim}{\text{j}}+5 \underset{\sim}{\text{k}}\), where \(p\) and \(q\) are real numbers.

If these vectors are linearly dependent, then

  1. \(8p=5q-35\)
  2. \(5p=8q-35\)
  3. \(8p=-5q-35\)
  4. \(8p=5q+35\)
  5. \(5p=8q+35\)
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Linearly dependent when}\ \underset{\sim}{c}=m\underset{\sim}{a}+n\underset{\sim}{b}\)

\(\underset{\sim}{c}=-3 \underset{\sim}{i}+2 \underset{\sim}{j}+\underset{\sim}{k}\)

\(m\underset{\sim}{a}+n \underset{\sim}{b}=(2 m+n) \underset{\sim}{i}-(3 m-2 n) \underset{\sim}{j}+(m p-q n) \underset{\sim}{k}\)

\(\text{Equating co-efficients:}\)

  \(2 m+n=-3 \ \ldots\ (1)\) 

  \(-3 m+2 n=2\ \ldots\ (2)\)

  \(p m-q n=5\ \ldots\ (3)\)

\(\text{Solve (1) and (2) by CAS:}\)

  \(m=-\dfrac{8}{7}, n=-\dfrac{5}{7}\)

\(\text { Substitute } m,n \text { into (3): }\)

\begin{aligned}
-\dfrac{8}{7}\,p+\dfrac{5}{7}\,q & =5 \\
8 p & =5 q-35
\end{aligned}

\(\Rightarrow A\)

PHYSICS, M5 2020 VCE 8 MC

A ball is attached to the end of a string and rotated in a circle at a constant speed in a vertical plane, as shown in the diagram below.
 

The arrows in options \(\text{A}\). to \(\text{D}\). below indicate the direction and the size of the forces acting on the ball.

Ignoring air resistance, which one of the following best represents the forces acting on the ball when it is at the bottom of the circular path and moving to the left?
 

Show Answers Only

\(D\)

Show Worked Solution

→ There are only two forces acting on the ball, gravitational force directly down, and the centripetal force (string tension) directly towards the centre of the circular motion.

→ As the ball is following a circular path, the tension in the string is greater than the gravitational force acting on the ball.

\(\Rightarrow D\)

Mean mark 56%.

PHYSICS, M6 2020 VCE 5 MC

A coil consisting of 20 loops with an area of 10 cm\(^2\) is placed in a uniform magnetic field \(B\) of strength 0.03 T so that the plane of the coil is perpendicular to the field direction, as shown in the diagram below.
 

The magnetic flux through the coil is closest to

  1. 0 Wb
  2. 3.0 × 10\(^{-5}\) Wb
  3. 6.0 × 10\(^{-4}\) Wb
  4. 3.0 × 10\(^{-1}\) Wb
Show Answers Only

\(B\)

Show Worked Solution

→ Convert cm\(^2\) to m\(^2\): 1 cm\(^2\) = 0.01 m × 0.01 m = 0.0001 m\(^2\)

→ 10 cm\(^2\) = 0.0010 m\(^2\)

→ \(\Phi = BA = 0.03 \times 0.0010 = 3.0 \times 10^{-5}\ \text{Wb}\)

\(\Rightarrow B\)

PHYSICS, M6 2020 VCE 3-4 MC

A positron with a velocity of 1.4 × 10\(^6\) m s\(^{-1}\) is injected into a uniform magnetic field of 4.0 × 10\(^{-2}\) T, directed into the page, as shown in the diagram below.

It moves in a vacuum in a semicircle of radius \(r\). The mass of the positron is 9.1 × 10\(^{-31}\) kg and the charge on the positron is 1.6 × 10\(^{-19}\) C. Ignore relativistic effects.
 


 

Question 3

Which one of the following best gives the speed of the positron as it exits the magnetic field?

  1. 0 m s\(^{-1}\)
  2. much less than 1.4 × 10\(^6\) m s\(^{-1}\)
  3. 1.4 × 10\(^6\) m s\(^{-1}\)
  4. greater than 1.4 × 10\(^6\) m s\(^{-1}\)

 
Question 4

The speed of the positron is changed to 7.0 × 10\(^5\) m s\(^{-1}\).

Which one of the following best gives the value of the radius \(r\) for this speed?

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

\(\text{Question 3:}\ C\)

\(\text{Question 4:}\ B\)

Show Worked Solution

\(\text{Question 3}\)

→ As the direction of the force is perpendicular to the velocity, the velocity will not change.

→ The velocity will be \(1.4 \times 10^6\ \text{ms}^{-1}\)

\(\Rightarrow C\)
 

\(\text{Question 4}\)

\(F_B\) \(=F_c\)  
\(qvB\) \(=\dfrac{mv^2}{r}\)  
\(r\) \(=\dfrac{mv}{qB}\)  

 
→ Therefore, \(r \propto v\).

→ If the velocity is halved, the radius will also be halved.

\(\Rightarrow B\)

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

Calculus, 2ADV C4 2023 MET1 5

  1. Evaluate  \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\).   (1 mark)
  2. Hence, or otherwise, find all values of \(k\) such that \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx=\displaystyle \int_{k}^{\frac{\pi}{2}} \cos(x)\,dx\), where \(-3\pi<k<2\pi\).   (3 marks)
Show Answers Only

a.    \(\dfrac{1}{2}\)

b.    \(k=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

Show Worked Solution
a.    \(\displaystyle \int_{0}^{\frac{\pi}{3}} \sin(x)\,dx\) \(=\left[-\cos x\right]_0^\frac{\pi}{3}\)
    \(=-\cos\dfrac{\pi}{3}+\cos 0\)
    \(=-\dfrac{1}{2}+1\)
    \(=\dfrac{1}{2}\)

 

b.    \(\displaystyle \int_{k}^{\frac{\pi}{2}} \cos(x)\,dx\) \(=\left[\sin x\right]_k^\frac{\pi}{2}\)
    \(=\sin\bigg(\dfrac{\pi}{2}\bigg)-\sin (k)\)
    \(=1-\sin (k)\)

 
\(\text{Using part (a):}\)

\(1-\sin (k)\) \(=\dfrac{1}{2}\)
\(\sin (k)\) \(=\dfrac{1}{2}\)
\(\therefore\ k\) \(=\dfrac{-11\pi}{6},\ \dfrac{-7\pi}{6},\ \dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)

PHYSICS, M7 2021 VCE 15

A photoelectric experiment is carried out by students. They measure the threshold frequency of light required for photoemission to be 6.5 × 10\(^{14}\) Hz and the work function to be 3.2 × 10\(^{-19}\) J.

Using the students' measurements, what value would they calculate for Planck's constant? Outline your reasoning and show all your working. Give your answer in joule-seconds.   (3 marks)

Show Answers Only

\(h=4.9 \times 10^{-34}\ \text{J s}\)

Show Worked Solution

\(f=6.5 \times 10^{14}\ \text{Hz},\ \ E=3.2 \times 10^{-19}\ \text{J}\)

\(E=hf\ \ \Rightarrow\ \ h=\dfrac{E}{f}\)

\(\therefore h=\dfrac{3.2 \times 10^{-19}}{6.5 \times 10^{14}} = 4.92 \times 10^{-34}\ \text{J s}\)

Calculus, MET2 2022 VCAA 2

On a remote island, there are only two species of animals: foxes and rabbits. The foxes are the predators and the rabbits are their prey.

The populations of foxes and rabbits increase and decrease in a periodic pattern, with the period of both populations being the same, as shown in the graph below, for all `t \geq 0`, where time `t` is measured in weeks.

One point of minimum fox population, (20, 700), and one point of maximum fox population, (100, 2500), are also shown on the graph.

The graph has been drawn to scale.
 

The population of rabbits can be modelled by the rule `r(t)=1700 \sin \left(\frac{\pi t}{80}\right)+2500`.

  1.   i. State the initial population of rabbits.   (1 mark)

  2.  ii. State the minimum and maximum population of rabbits.   (1 mark)

  3. iii. State the number of weeks between maximum populations of rabbits.   (1 mark)

The population of foxes can be modelled by the rule `f(t)=a \sin (b(t-60))+1600`.

  1. Show that `a=900` and `b=\frac{\pi}{80}`.   (2 marks)

  2. Find the maximum combined population of foxes and rabbits. Give your answer correct to the nearest whole number.   (1 mark)

  3. What is the number of weeks between the periods when the combined population of foxes and rabbits is a maximum?   (1 mark)

The population of foxes is better modelled by the transformation of `y=\sin (t)` under `Q` given by
 

  1. Find the average population during the first 300 weeks for the combined population of foxes and rabbits, where the population of foxes is modelled by the transformation of `y=\sin(t)` under the transformation `Q`. Give your answer correct to the nearest whole number.   (4 marks)

Over a longer period of time, it is found that the increase and decrease in the population of rabbits gets smaller and smaller.

The population of rabbits over a longer period of time can be modelled by the rule

`s(t)=1700cdote^(-0.003t)cdot sin((pit)/80)+2500,\qquad text(for all)\ t>=0`

  1. Find the average rate of change between the first two times when the population of rabbits is at a maximum. Give your answer correct to one decimal place.   (2 marks)

  2. Find the time, where `t>40`, in weeks, when the rate of change of the rabbit population is at its greatest positive value. Give your answer correct to the nearest whole number.   (2 marks)

  3. Over time, the rabbit population approaches a particular value.
  4. State this value.   (1 mark)

Show Answers Only

ai.   `r(0)=2500`

aii.  Minimum population of rabbits `= 800`

Maximum population of rabbits `= 4200`

aiii. `160`  weeks

b.    See worked solution.

c.    `~~ 5339` (nearest whole number)

d.    Weeks between the periods is 160

e.    `~~ 4142` (nearest whole number)

f.    Average rate of change `=-3.6` rabbits/week (1 d.p.)

g.    `t = 156` weeks (nearest whole number)  

h.    ` s → 2500`

Show Worked Solution

ai.  Initial population of rabbits

From graph when `t=0, \ r(0) = 2500`

Using formula when `t=0`

`r(t)` `= 1700\ sin \left(\frac{\pi t}{80}\right)+2500`  
`r(0)` `= 1700\ sin \left(\frac{\pi xx 0}{80}\right)+2500 = 2500`  rabbits  


aii. From graph,

Minimum population of rabbits `= 800`

Maximum population of rabbits `= 4200`

OR

Using formula

Minimum is when `t = 120`

`r(120) = 1700\ sin \left(\frac{\pi xx 120}{80}\right)+2500 = 1700 xx (-1) + 2500 = 800`

Maximum is when `t = 40`

`r(40) = 1700\ sin \left(\frac{\pi xx 40}{80}\right)+2500 = 1700 xx (1) + 2500 = 4200`

 
aiii. Number of weeks between maximum populations of rabbits `= 200-40 = 160`  weeks

 
b.  Period of foxes = period of rabbits = 160:

`frac{\2pi}{b} = 160`

`:.\  b = frac{\2pi}{160} = frac{\pi}{80}` which is the same period as the rabbit population.

Using the point `(100 , 2500)`

Amplitude when `b = frac{\pi}{80}`: 

`f(t)` `=a \ sin (pi/80(t-60))+1600`  
`f(100)` `= 2500`  
`2500` `= a \ sin (pi/80(100-60))+1600`  
`2500` `= a \ sin (pi/2)+1600`  
`a` `= 2500-1600 = 900`  

 
`:.\  f(t)= 900 \ sin (pi/80)(t-60) + 1600`

  

c.    Using CAS find `h(t) = f(t) + r(t)`:

`h(t):=900 \cdot \sin \left(\frac{\pi}{80} \cdot(t-60)\right)+1600+1700\cdot \sin \left(\frac{\pit}{80}\right) +2500`

  
`text{fMax}(h(t),t)|0 <= t <= 160`      `t = 53.7306….`
  

`h(53.7306…)=5339.46`

  
Maximum combined population `~~ 5339` (nearest whole number)


♦♦ Mean mark (c) 40%.
MARKER’S COMMENT: Many rounding errors with a common error being 5340. Many students incorrectly added the max value of rabbits to the max value of foxes, however, these points occurred at different times.

d.    Using CAS, check by changing domain to 0 to 320.

`text{fMax}(h(t),t)|0 <= t <= 320`     `t = 213.7305…`
 
`h(213.7305…)=5339.4568….`
 
Therefore, the number of weeks between the periods is 160.
  

e.    Fox population:

`t^{\prime} = frac{90}{pi}t + 60`   →   `t = frac{pi}{90}(t^{\prime}-60)`

`y^{\prime} = 900y+1600`   →   `y = frac{1}{900}(y^{\prime}-1600)`

`frac{y^{\prime}-1600}{900} = sin(frac{pi(t^{\prime}-60)}{90})`

`:.\  f(t) = 900\ sin\frac{pi}{90}(t-60) + 1600`

  
Average combined population  [Using CAS]   
  
`=\frac{1}{300} \int_0^{300} left(\900 \sin \left(\frac{\pi(t-60)}{90}\right)+1600+1700\ sin\ left(\frac{\pi t}{80}\right)+2500\right) d t`
  

`= 4142.2646…..  ~~ 4142` (nearest whole number)


♦ Mean mark (e) 40%.
MARKER’S COMMENT: Common incorrect answer 1600. Some incorrectly subtracted `r(t)`. Others used average rate of change instead of average value.

f.   Using CAS

`s(t):= 1700e^(-0.003t) dot\sin\frac{pit}{80} + 2500`
 

`text{fMax}(s(t),t)|0<=t<=320`        `x = 38.0584….`
 

`s(38.0584….)=4012.1666….`
 

`text{fMax}(s(t),t)|160<=t<=320`     `x = 198.0584….`
 

`s(198.0584….)=3435.7035….`
 

Av rate of change between the points

`(38.058 , 4012.167)`  and  `(198.058 , 3435.704)`

`= frac{4012.1666….-3435.7035….}{38.0584….-198.0584….} =-3.60289….`
 

`:.` Average rate of change `=-3.6` rabbits/week (1 d.p.)


♦ Mean mark (f ) 45%.
MARKER’S COMMENT: Some students rounded too early.
`frac{s(200)-s(40)}{200-40}` was commonly seen.
Some found average rate of change between max and min populations.

g.   Using CAS

`s^(primeprime)(t) = 0` , `t = 80(n-0.049) \ \forall n \in Z`

After testing `n = 1, 2, 3, 4` greatest positive value occurs for `n = 2` 

`t` `= 80(n-0.049)`  
  `= 80(2-0.049)`  
  `= 156.08`  

 
`:. \ t = 156` weeks (nearest whole number)


♦♦♦ Mean mark (g) 25%.
MARKER’S COMMENT: Many students solved `frac{ds}{dt}=0`.
A common answer was 41.8, as `s(156.11…)=41.79`.
Another common incorrect answer was 76 weeks.

h.   As `t → ∞`, `e^(-0.003t) → 0`

`:.\ s → 2500`

Calculus, MET2 2022 VCAA 1

The diagram below shows part of the graph of `y=f(x)`, where `f(x)=\frac{x^2}{12}`.
 

  1. State the equation of the axis of symmetry of the graph of `f`.   (1 mark)

  2. State the derivative of `f` with respect to `x`.   (1 mark)

The tangent to `f` at point `M` has gradient `-2` .

  1. Find the equation of the tangent to `f` at point `M`.   (2 marks)

The diagram below shows part of the graph of `y=f(x)`, the tangent to `f` at point `M` and the line perpendicular to the tangent at point `M`.
 

 

  1.  i. Find the equation of the line perpendicular to the tangent passing through point `M`.   (1 mark)

  2. ii. The line perpendicular to the tangent at point `M` also cuts `f` at point `N`, as shown in the diagram above.
  3.     Find the area enclosed by this line and the curve `y=f(x)`.   (2 marks)

  4. Another parabola is defined by the rule `g(x)=\frac{x^2}{4 a^2}`, where `a>0`.
  5. A tangent to `g` and the line perpendicular to the tangent at `x=-b`, where `b>0`, are shown below.

  1. Find the value of `b`, in terms of `a`, such that the shaded area is a minimum.   (4 marks)

Show Answers Only

a.    `x=0`

b.    ` f^{\prime}(x)=1/6x`

c.    `x=-12`

di.    `y=1/2x + 18`

dii.   Area`= 375` units²

e.    `b = 2a^2`

Show Worked Solution

a.   Axis of symmetry:  `x=0`
  

b.    `f(x)`  `=\frac{x^2}{12}`  
  ` f^{\prime}(x)` `= 1/6x`  

  
c.   At `M` gradient `= -2`

`1/6x` `= -2`  
`x` `= -12`  

 
When  `x = -12`
 

`f(x) = (-12)^2/12 = 12`
 
Equation of tangent at `(-12 , 12)`:

`y-y_1` `=m(x-x_1)`  
`y-12` `= -2(x + 12)`  
`y` `= -2x-12`  


d.i   Gradient of tangent `= -2`

`:.`  gradient of normal `= 1/2` 

Equation at `M(- 12 , 12)`

`y -y_1` `=m(x-x_1)`  
`y-12` `= 1/2(x + 12)`  
`y` `=1/2x + 18`  


d.ii  Points of intersection of `f(x)` and normal are at  `M` and `N`.

So equate ` y = x^2/12` and  `y = 1/2x + 18` to find `N`

`x^2/12` `=1/2x + 18`  
`x^2-6x-216` `=0`  
`(x + 12)(x-18)` `=0`  

   
`:.\  x = -12` or `x = 18`

Area `= \int_{-12}^{18}\left(\frac{1}{2} x+18-\frac{x^2}{12}\right) d x`  
  `= [x^2/4 + 18x-x^3/36]_(-12)^18`  
  `= [18^2/4 +18^2-18^3/36] – [12^2/4 + 18 xx (-12)-(-12)^3/36]`  
  `= 375` units²  

 

e.   `g(x) = x^2/(4a^2)`   `a > 0`

At `x = -b`   `y = (-b)^2/(4a^2) = b^2/4a^2`

`g^{\prime}(x) = (2x)/(4a^2) = x/(2a^2)`

Gradient of tangent `= (-b)/(2a^2)`

Gradient of normal `= (2a^2)/b`

Equation of normal at `(- b , b^2/(4a^2))`

`y-y_1` `= m(x-x_1)`  
`y-b^2/(4a^2)` `= (2a^2)/b(x-(-b))`  
`y` `= (2a^2x)/b + 2a^2 + b^2/(4a^2)`  
`y` `= (2a^2x)/b +(8a^4 + b^2)/(4a^2)`  

 
Points of intersection of normal and parabola (Using CAS)

solve `((2a^2x)/b +(8a^4 + b^2)/(4a^2) = x^2/(4a^2),x)`

`x =-b`  or  `x = (8a^4+b^2)/b`

 
Calculate area using CAS

`A = \int_{-b}^{(8a^4+b^2)/b}\left(\frac{2a^2x}{b} +\frac{8a^4 + b^2}{4a^2}-frac{x^2}{4a^2} \right) dx`

`A = frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3}`
 
Using CAS Solve derivative of `A = 0`  with respect to `b` to find `b`

solve`(d/(db)(frac{64a^12 + 48a^8b^2 + 12a^4b^4 + b^6}{3a^2b^3})=0,b)`

 
`b =-2a^2`   and  `b = 2a^2`

Given `b > 0`

`b = 2a^2`


♦♦ Mean mark (e) 33%.
MARKER’S COMMENT: Common errors: Students found `\int_{-b}^{\frac{8 a^4+b^2}{b}}\left(y_n\right) d x` and failed to subtract `g(x)` or had incorrect terminals.

Calculus, MET2 2022 VCAA 16 MC

The function `f(x)=\frac{1}{3} x^3+m x^2+n x+p`, for `m, n, p \in R`, has turning points at `x=-3` and `x=1` and passes through the point (3, 4).

The values of `m, n` and `p` respectively are

  1. `m=0, \quad n=-\frac{7}{3}, p=2`
  2. `m=1, n=-3, \quad p=-5`
  3. `m=-1, n=-3, \quad p=13`
  4. `m=\frac{5}{4}, \quad n=\frac{3}{2}, \quad p=-\frac{83}{4}`
  5. `m=\frac{5}{2}, \quad n=6, \quad p=-\frac{91}{2}`
Show Answers Only

`B`

Show Worked Solution

For turning point `f^{\prime}(x) = 0`

`f(x)` `=\frac{1}{3} x^3 + mx^2 + nx + p`  
`:.\ f^{\prime}(x)` `= x^2-2mx+n`  
` f^{\prime}(- 3)` `= 9 – 6m + n`   …. (1)  
` f^{\prime}( 1)` `= 1 + 2m + n`   ….(2)  

  

Solving (1) and (2)     `9  –  6m + n`  `= 0`
  `1 + 2m + n` `= 0`

`m = 1`  and `n = – 3`

`:. \ f(x)=\frac{1}{3} x^3 + x^2- 3x + p`
  

The point `(3 , 4)` lies on the line

`4` `= \frac{1}{3} xx 3^3 +3^2- 3 xx 3 + p`  
`:. \ p` `= – 5 `  

 
`:. \ f(x)=\frac{1}{3} x^3 + x^2- 3x – 5`

→  `m=1, n=-3, \quad p=-5`
  
 `=>B`

 

Probability, MET2 2022 VCAA 14 MC

A continuous random variable, \(X\), has a probability density function given by
  

\(f(x)=\begin{cases}\dfrac{2}{9}xe^{-\frac{1}{9}x^2}         &\ \ x\geq 0 \\ \\ 0       &\ \ x<0 \\ \end{cases}\)
 

The expected value of \(X\), correct to three decimal places, is

  1. 1.000
  2. 2.659
  3. 3.730
  4. 6.341
  5. 9.000
Show Answers Only

\(B\)

Show Worked Solution
\(\text{E}(X) = \displaystyle\int_0^{\infty} \dfrac{2}{9} x^2 e^{-\frac{1}{9} x^2}\,dx\) \(\approx 2.65868\dots\)    [by CAS]  
   \(\approx 2.659\)  

  
\(\Rightarrow B\)

Calculus, MET2 2022 VCAA 11 MC

If `\frac{d}{d x}(x \cdot \sin(x))=\sin (x)+x \cdot \cos(x)`, then `\frac{1}{k} \int x \ cos(x)dx` is equal to

  1. `k\left(x \cdot \sin (x)-\int \sin (x) d x\right)+c`
  2. `\frac{1}{k} x \cdot \sin (x)-\int \sin (x) d x+c`
  3. `\frac{1}{k}\left(x \cdot \sin (x)-\int \sin (x) d x\right)+c`
  4. `\frac{1}{k}(x \cdot \sin (x)-\sin (x))+c`
  5. `\frac{1}{k}\left(\int x \cdot \sin (x) d x-\int \sin (x) d x\right)+c`
Show Answers Only

`C`

Show Worked Solution

Given `\frac{d}{d x}(x \cdot \sin(x))=\sin (x)+x \cdot \cos(x)`, then

`x\cdot \cos (x)` `=\frac{d}{d x}(x\cdot \sin (x))-\sin (x)`  
`\frac{1}{k} \int x \cos (x) d x` `= \frac{1}{k}\left(\int \frac{d}{d x}(x \cdot \sin (x)) d x-\int \sin (x) d x\right)`  
  `= \frac{1}{k}\left(x \cdot \sin (x)-\int \sin (x)d x\right)+c`  

 
`=> C`