SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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)

Show Answers Only

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 12

In a Young's double-slit interference experiment, laser light is incident on two slits, \(\text{S}_1\) and \(\text{S}_2\), that are 4.0 × 10\(^{-4}\) m apart, as shown in Figure 11a.

Rays from the slits meet on a screen 2.00 m from the slits to produce an interference pattern. Point \(\text{C}\) is at the centre of the pattern. Figure 11b shows the pattern obtained on the screen.
 

  1. There is a bright fringe at point \(\text{P}\) on the screen.
  2. Explain how this bright fringe is formed.  (2 marks)
  1. The distance from the central bright fringe at point \(\text{C}\) to the bright fringe at point P is 1.26 × 10\(^{-2}\) m.
  2. Calculate the wavelength of the laser light. Show your working.  (3 marks

Show Answers Only

a.    Bright fringe at point \(\text{P}\):

→ Young’s double slit experiment demonstrates that light can produce an interference pattern, from both constructive and destructive interference.

→ Point \(\text{P}\) is the 4th bright fringe and it follows that the path difference between the two light beams is 4 wavelengths.

→ At this point, the peaks of the twos wave constructively interfere and a bright band is formed.

b.    \(630\ \text{nm}\)

Show Worked Solution

a.    Bright fringe at point \(\text{P}\):

→ Young’s double slit experiment demonstrates that light can produce an interference pattern, from both constructive and destructive interference.

→ Point \(\text{P}\) is the 4th bright fringe and it follows that the path difference between the two light beams is 4 wavelengths.

→ At this point, the peaks of the twos wave constructively interfere and a bright band is formed.
 

♦ Mean mark (a) 45%.

b.    Using  \(d\,\sin \theta=m \lambda\)  and  \(\sin \theta=\dfrac{\Delta x}{D}\)

→ \(\Delta x\) is the distance between two bright bands and \(D\) is the distance from the slits to the screen.

\(\dfrac{d\Delta x}{D}\) \(=m\lambda\)  
\(\lambda\) \(=\dfrac{d\Delta x}{Dm}\)  
  \(=\dfrac{4 \times 10^{-4} \times 1.26 \times 10^{-2}}{2 \times 4}\)  
  \(=630\ \text{nm}\)  
♦ Mean mark (b) 40%.

PHYSICS, M7 2020 VCE 11

An astronaut has left Earth and is travelling on a spaceship at 0.800\(c\) directly towards the star known as Sirius, which is located 8.61 light-years away from Earth, as measured by observers on Earth.

  1. How long will the trip take according to a clock that the astronaut is carrying on his spaceship? Show your working.   (2 marks)
  1. Is the trip time measured by the astronaut in part a. a proper time? Explain your reasoning.   (2 marks)

Show Answers Only

a.    \(6.46\ \text{years}\)

b.   → The trip time of 6.46 years on the spaceship is a proper time.

→ This is due to the Astronaut’s clock being stationary within the astronaut’s frame of reference.

Show Worked Solution

a.    From the earth’s perspective:

\(\text{Travel time}\ =\dfrac{8.61}{0.8}=10.76\ \text{years}\).
 

From the astronaut’s perspective:

→  The Earth’s time is going to be dilated compared to his, so the time the astronauts clock will measure is:

\(t=\dfrac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\)

\(t_0\) \(=t\sqrt{1-\frac{v^2}{c^2}}\)
  \(=10.76 \times \sqrt{1-\frac{(0.8c)^2}{c^2}}\)
  \(=10.76 \times \sqrt{1-0.8^2}\)
  \(=6.46\ \text{years}\)

♦♦♦ Mean mark (a) 29%.
COMMENT: Light years is a measure of distance, not time!

b.   → The trip time of 6.46 years on the spaceship is a proper time.

→ This is due to the Astronaut’s clock being stationary within the astronaut’s frame of reference.

♦ Mean mark (b) 44%.

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)
Show Answers Only

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.

Graphs, MET1 2023 VCE SM-Bank 1

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

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

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

Show Answers Only

a.
   

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

Show Worked Solution

a.
     

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

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

Calculus, MET2 2023 VCE SM-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.

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 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 3

Carbon dioxide is produced during the combustion of propane \(\ce{(C3H8)}\) in oxygen \(\ce{(O2)}\). The balanced chemical equation for this reaction is:

\(\ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)}\)

If 44.0 grams of propane are completely combusted, calculate the volume of carbon dioxide produced at STP (100 kPa and 0\(^{\circ}\)C).    (3 marks)

Show Answers Only

\(\ce{V(CO2)}\ =68\ \text{L}\)

Show Worked Solution

Calculate the number of moles of propane (\(\ce{C3H8}\)):

  \(\ce{n(C3H8) = \dfrac{\text{m}}{\text{MM}} = \dfrac{44.0\ \text{g}}{44.094\ \text{g mol}^{-1}} = 0.998\ \text{mol}}\)
 

 Use the stoichiometric ratio to find the moles of \(\ce{CO2}\) produced:

  \(\ce{n(CO2) = 3 \times n(C3H8) = 3 \times 0.998 \, mol = 2.994 \, mol}\)
 

Calculate the volume of \(\ce{CO2}\) at STP:

  \(\ce{V(CO2) = n \times 22.71 \, L \, mol^{-1} = 2.994 \, mol \times 22.71 \, L \, mol^{-1} = 68 \, L}\)

→ The volume of carbon dioxide produced is 68 litres.

Calculus, MET2 2023 VCE SM-Bank 6 MC

Consider the algorithm below, which uses the bisection method to estimate the solution to an equation in the form \(f(x)=0\).

The algorithm is implemented as follows.
 

Which value would be returned when the algorithm is implemented as given?

  1. -0.351
  2. -0.108
  3. 3.25
  4. 3.5
  5. 4
Show Answers Only

\(D\)

Show Worked Solution

\(\textbf{Define }\text{bisection}\ (f(x),a,b,\max)\ \rightarrow\ (\sin(x), 3, 5, 2)\)

\(\text{Test for }\ i=0, 1\ \rightarrow i<\max=2\)

\(i=0\quad\) \(\text{mid}\) \(=\dfrac{3+5}{2}\) \(=4\quad [\sin(4)<0\ \text{ and }\ \sin(3)\times\sin(4)<0]\)
\(i=1\quad\) \(\text{mid}\) \(=\dfrac{3+4}{2}\) \(=3.5\quad [\sin(3.5)<0\ \text{ and }\ \sin(3)\times\sin(3.5)<0]\)

    

\(\Rightarrow D\)

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    \(g(x)\) \(=3\log_{e}x-x\)
  \(g(1)\) \(=3\log_{e}1-1=-1\)
  \(g'(x)\) \(=\dfrac{3}{x}-1\)
  \(g'(1)\) \(=\dfrac{3}{1}-1=2\)

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

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

b.ii

 
c.    \(\text{Newton’s Method}\)

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

\(\text{Using CAS:}\)
  
 

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, 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.

CHEMISTRY, M1 EQ-Bank 35

Explain why ethanol \(\ce{(C2H5OH)}\) is a liquid at room temperature whereas ethane \(\ce{(C2H6)}\) is a gas.    (3 marks)

Show Answers Only

→ Ethanol \(\ce{(C2H5OH)}\) is a polar molecule with an \(\ce{(-OH)}\) group that can form hydrogen bonds.

→ Hydrogen bonds are strong intermolecular forces that require more energy to break, resulting in ethanol being a liquid at room temperature.

→ Ethane \(\ce{(C2H6)}\), however, is a nonpolar molecule. The only intermolecular forces present in ethane are dispersion forces (also known as London dispersion forces), which are much weaker than hydrogen bonds.

→ As a result, ethane remains a gas at room temperature because less energy is required to separate its molecules.

→ The significant difference in the strength of intermolecular forces (hydrogen bonding in ethanol vs. dispersion forces in ethane) explains why ethanol is a liquid and ethane is a gas at room temperature.

Show Worked Solution

→ Ethanol \(\ce{(C2H5OH)}\) is a polar molecule with an \(\ce{(-OH)}\) group that can form hydrogen bonds.

→ Hydrogen bonds are strong intermolecular forces that require more energy to break, resulting in ethanol being a liquid at room temperature.

→ Ethane \(\ce{(C2H6)}\), however, is a nonpolar molecule. The only intermolecular forces present in ethane are dispersion forces (also known as London dispersion forces), which are much weaker than hydrogen bonds.

→ As a result, ethane remains a gas at room temperature because less energy is required to separate its molecules.

→ The significant difference in the strength of intermolecular forces (hydrogen bonding in ethanol vs. dispersion forces in ethane) explains why ethanol is a liquid and ethane is a gas at room temperature.

CHEMISTRY, M1 EQ-Bank 34

Carbon exhibits allotropy, meaning it can exist in different forms with distinct physical properties.

Describe two carbon allotropes, graphite and diamond, and explain how their structural differences result in their distinct physical properties.   (4 marks)

Show Answers Only

Graphite:

→ Graphite consists of layers of carbon atoms arranged in a hexagonal lattice.

→ Each carbon atom is bonded to three others in the same plane, with delocalized electrons between layers.

→ The weak intermolecular forces between these layers allow them to slide over each other easily, contributing to graphite’s softness and its use as a lubricant.
 

Diamond:

→ In contrast, diamond features a three-dimensional lattice where each carbon atom forms four strong covalent bonds with other carbon atoms, arranged in a tetrahedral structure.

→ This extensive network of strong bonds throughout the lattice makes diamond one of the hardest natural substances, leading to its use in cutting and drilling tools.

Show Worked Solution

Graphite:

→ Graphite consists of layers of carbon atoms arranged in a hexagonal lattice.

→ Each carbon atom is bonded to three others in the same plane, with delocalized electrons between layers.

→ The weak intermolecular forces between these layers allow them to slide over each other easily, contributing to graphite’s softness and its use as a lubricant.
 

Diamond:

→ In contrast, diamond features a three-dimensional lattice where each carbon atom forms four strong covalent bonds with other carbon atoms, arranged in a tetrahedral structure.

→ This extensive network of strong bonds throughout the lattice makes diamond one of the hardest natural substances, leading to its use in cutting and drilling tools.

CHEMISTRY, M1 EQ-Bank 31

Using your knowledge of electronic configurations, explain what happens to atomic radii as you go across a period from left to right in the periodic table.

Identify one element from the first period with a larger atomic radius and one with a smaller atomic radius than Boron.   (3 marks)

Show Answers Only

→ As elements progress from left to right across a period, each successive element has one more proton and electron than the previous one.

→ These additional electrons enter the same energy level, increasing the effective nuclear charge experienced by electrons.

→ However, the stronger pull from the nucleus due to the higher positive charge causes the electrons to be drawn closer, resulting in a decrease in atomic radius.

→ Lithium has a larger atomic radius than Boron, while Neon has a smaller atomic radius than Boron.

Show Worked Solution

→ As elements progress from left to right across a period, each successive element has one more proton and electron than the previous one.

→ These additional electrons enter the same energy level, increasing the effective nuclear charge experienced by electrons.

→ However, the stronger pull from the nucleus due to the higher positive charge causes the electrons to be drawn closer, resulting in a decrease in atomic radius.

→ Lithium has a larger atomic radius than Boron, while Neon has a smaller atomic radius than Boron.

Calculus, MET2 2023 VCE SM-Bank 7 MC

One way of implementing Newton's method using pseudocode, with a tolerance level of 0.001 , is shown below.

The pseudocode is incomplete, with two missing lines indicated by an empty box.
 

  

Which one of the following options would be most appropriate to fill the empty box?
 



Show Answers Only

\(E\)

Show Worked Solution

\(\text{The tolerance}=\pm 0.001\)

\(\Big|\text{next_ x}-\text{prev_ x}\Big|<\ \text{tolerance}\)

\(\therefore\ \textbf{If }\ \ -0.001<\ \text{next_ x}-\text{prev_ x}\ <0.001\ \textbf{Then }\)

\(\text{If true }\textbf{Return }\text{next_ x}\)

\(\Rightarrow E\)

CHEMISTRY, M1 EQ-Bank 28

Using your understanding of periodic trends, explain and predict the differences in the properties of the elements lithium (Li) and fluorine (F) regarding their:

    • atomic radii
    • first ionisation energy
    • electronegativity

Justify your predictions based on their positions in the periodic table and electronic configurations.   (4 marks)

Show Answers Only

Atomic Radii:

→ Lithium, being in the same period but a different group than fluorine, has fewer protons and a less effective nuclear charge, resulting in a larger atomic radius.

First Ionisation Energy:

→ Fluorine has a higher first ionisation energy due to its greater nuclear charge and smaller radius, which strongly attracts electrons.

Electronegativity:

→ Fluorine, being one of the most electronegative elements, has a strong ability to attract bonding electrons.

Show Worked Solution

Atomic Radii:

→ Lithium, being in the same period but a different group than fluorine, has fewer protons and a less effective nuclear charge, resulting in a larger atomic radius.

First Ionisation Energy:

→ Fluorine has a higher first ionisation energy due to its greater nuclear charge and smaller radius, which strongly attracts electrons.

Electronegativity:

→ Fluorine, being one of the most electronegative elements, has a strong ability to attract bonding electrons.

Calculus, MET2 2023 VCE SM-Bank 5 MC

The algorithm below, described in pseudocode, estimates the value of a definite integral using the trapezium rule.
 

Consider the algorithm implemented with the following inputs.

The value of the variable sum after one iteration of the while loop would be closest to

  1. 1.281
  2. 1.289
  3. 1.463
  4. 1.617
  5. 2.136
Show Answers Only

\(C\)

Show Worked Solution

\((f(x), a, b, n)\ \rightarrow\ (\log_{e}x, 1, 3, 10) \)

\(h=\dfrac{b-a}{n}=\dfrac{3-1}{10}=\dfrac{1}{5}\)

\(\text{Sum}=f(a)+f(b)=\log_{e}1+\log_{e}3=\log_{e}3\)
  

\(\text{1st iteration of}\ \textbf{while }\text{loop:}\)

\(x\) \(=a+h=1+\dfrac{1}{5}=\dfrac{6}{5}\)
\(\text{Sum}\) \(=\text{Sum}+2\times f(x)\)
  \(=\log_{e}3+2\times f\Bigg(\dfrac{6}{5}\Bigg)\)
  \(=\log_{e}3+2\times \log_{e}{\Bigg(\dfrac{6}{5}\Bigg)}\)
  \(\approx 1.46325\dots\)

 

\(\Rightarrow C\)

Calculus, MET2 2023 VCE SM-Bank 2 MC

Newton's method is being used to approximate the non-zero \(x\)-intercept of the function with the equation  \(f(x)=\dfrac{x^3}{5}-\sqrt{x}\).  An initial estimate of  \(x_0=1\) is used.

Which one of the following gives the first estimate that would correctly approximate the intercept to three decimal places?

  1. \(x_6\)
  2. \(x_7\)
  3. \(x_8\)
  4. \(x_9\)
  5. The intercept cannot be correctly approximated using Newton's method.
Show Answers Only

\(C\)

Show Worked Solution
\(f(x)\) \(=\dfrac{x^3}{5}-\sqrt{x}\)
\(f'(x)\) \(=\dfrac{3x^2}{5}-\dfrac{1}{2\sqrt{x}}\)

\(\text{For Newton’s Method: }\)

\(x_o\) \(=1\)
\(x_{n+1}\) \(=x_n-\dfrac{f(x_n)}{f'(x_n)}\)

  

\(\text{Check using CAS:}\)

\begin{array} {|c|c|}
\hline
\ \ \ n\ \  \ & x_n \\
\hline
\ 0 \ & 1\\
\hline
\ 1 \ & 1-\dfrac{\dfrac{1^3}{5}-\sqrt{1}}{\dfrac{3\times 1^2}{5}-\dfrac{1}{2\sqrt{1}}}=9\\
\hline
\ 2 \ & 9-\dfrac{\dfrac{9^3}{5}-\sqrt{9}}{\dfrac{3\times 9^2}{5}-\dfrac{1}{2\sqrt{9}}}\approx 6.05\\
\hline
\ 3 \ & 6.05-\dfrac{\dfrac{6.05^3}{5}-\sqrt{6.05}}{\dfrac{3\times 6.05^2}{5}-\dfrac{1}{2\sqrt{6.05}}}\approx 4.13\\
\hline
\ 4 \ & 4.13-\dfrac{\dfrac{4.13^3}{5}-\sqrt{4.13}}{\dfrac{3\times 4.13^2}{5}-\dfrac{1}{2\sqrt{4.13}}}\approx 2.92\\
\hline
\ 5 \ & 2.92-\dfrac{\dfrac{2.92^3}{5}-\sqrt{2.92}}{\dfrac{3\times 2.92^2}{5}-\dfrac{1}{2\sqrt{2.92}}}\approx 2.24\\
\hline
\ 6 \ & 2.24-\dfrac{\dfrac{2.24^3}{5}-\sqrt{2.24}}{\dfrac{3\times 2.24^2}{5}-\dfrac{1}{2\sqrt{2.24}}}\approx 1.96\\
\hline
\ 7 \ & 1.96-\dfrac{\dfrac{1.96^3}{5}-\sqrt{1.96}}{\dfrac{3\times 1.96^2}{5}-\dfrac{1}{2\sqrt{1.96}}}\approx 1.906\\
\hline
\ 8 \ & 1.906-\dfrac{\dfrac{1.906^3}{5}-\sqrt{1.906}}{\dfrac{3\times 1.906^2}{5}-\dfrac{1}{2\sqrt{1.906}}}\approx 1.90366\\
\hline
\end{array}

  
\(\Rightarrow C\)

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}`.
 

  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.

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.   i. Find \(\text{Pr}(X=5)\).   (1 mark)

     ii. Find \(\text{Pr}(X \geq 2).\) (1 mark)

  2. iii. Find \(\text{Pr}(X \geq 2 | X<5)\), correct to three decimal places.   (2 marks)

  3.  iv. 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.   i. State the value of the definite integral \(\displaystyle\int_{1.5}^3 f(h)\,dh\).   (1 mark)

  2.  ii. 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. iii. 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)

  6. 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 .
  7. Bella flips her coin 25 times in order to estimate \(p\).
  8. 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`    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.

L&E, 2ADV E1 SM-Bank 15

Solve the following equation for \(x\):

\(2^{2x}=3(2^{x+1})-8\).   (3 marks)

Show Answers Only

\(x=1\ \ \text{or} \ \ 2\)

Show Worked Solution
\(2^{2x}\) \(=3(2^{x+1})-8\)  
\(2^{2x}\) \(=3(2 \times 2^{x})-8\)  
\(0\) \(=2^{2x}-6\cdot2^{x}+8\)  

 
\(\text{Let}\ \ X=2^{x}\)

\(X^2-6X+8\) \(=0\)  
\((X-4)(X-2)\) \(=0\)  
\(X\) \(=4\ \ \text{or}\ \ 2\)  

 
\(2^{x}=4\ \ \Rightarrow\ \ x=2\)

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

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%.

Statistics, SPEC2 2022 VCAA 18 MC

The time taken, \(T\) minutes, for a student to travel to school is normally distributed with a mean of 30 minutes and a standard deviation of 2.5 minutes.

Assuming that individual travel times are independent of each other, the probability, correct to four decimal places, that two consecutive travel times differ by more than 6 minutes is

  1. 0.0448
  2. 0.0897
  3. 0.1151
  4. 0.2301
  5. 0.9103
Show Answers Only

\(B\)

Show Worked Solution

\(T_d=T_1-T_2\)

\(E(T_d)=E(T_1)-E(T_2)=30-30-0\)

\(\text{Var}(T_1)=\ \text{Var}(T_2) = 2.5^2=6.25\)

\(\text{Var}(T_d)=1^2 \times \text{Var}(T_1)+(-1)^2 \times \text{Var}(T_2) = 12.5 \)

\(\sigma(T_d) = \sqrt{12.5}\)

\(1-\text{Pr}(-6 \lt T_d \lt 6)=0.0897 \ \ \text{(by CAS)}\)

\(\Rightarrow B\)

♦ Mean mark 42%.

Calculus, 2ADV C1 2023 MET2 11 MC

Two functions, \(f\) and \(g\), are continuous and differentiable for all  \(x\in R\). It is given that  \(f(-2)=-7,\ g(-2)=8\)  and  \(f^{′}(-2)=3,\ g^{′}(-2)=2\).

The gradient of the graph  \(y=f(x)\times g(x)\)  at the point where  \(x=-2\)  is

  1. \(-6\)
  2. \(0\)
  3. \(6\)
  4. \(10\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Using the Product Rule when}\ \ x=-2:\)

\(\dfrac{d}{dx}(f(x)\times g(x))\) \(=f(x)g^{′}(x)+g(x)f^{′}(x)\)
  \(=f(-2)g^{′}(-2)+g(-2)f^{′}(-2)\)
  \(=-7\times 2+8\times 3\)
  \(=10\)

 
\(\Rightarrow D\)


♦♦♦ Mean mark 22%.

Probability, 2ADV S1 2023 MET2 8 MC

A box contains \(n\) green balls and \(m\) red balls. A ball is selected at random, and its colour is noted. The ball is then replaced in the box.

In 8 such selections, where \(n\neq m\), what is the probability that a green ball is selected at least once?

  1. \(8\Bigg(\dfrac{n}{n+m}\Bigg)\Bigg(\dfrac{m}{n+m}\Bigg)^7\)
  2. \(1-\Bigg(\dfrac{n}{n+m}\Bigg)^8\)
  3. \(1-\Bigg(\dfrac{m}{n+m}\Bigg)^8\)
  4. \(1-\Bigg(\dfrac{n}{n+m}\Bigg)\Bigg(\dfrac{m}{n+m}\Bigg)^7\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{In any single selection:}\)

\(P(\text{green})\ =\Bigg(\dfrac{n}{n+m}\Bigg), \ \ P(\text{not green})\ =\Bigg(\dfrac{m}{n+m}\Bigg) \)

\(\text{Let}\ \ X=\ \text{choosing a green ball}\)

\(P(X\geq 1)\) \(=1-\text{Pr}(X=0)\)
  \(=1-\Bigg(\dfrac{m}{n+m}\Bigg)^8\)

 
\(\Rightarrow C\)


♦ Mean mark 49%.

PHYSICS, M5 2020 VCE 11 MC

The International Space Station (ISS) is travelling around Earth in a stable circular orbit, as shown in the diagram below.
 

Which one of the following statements concerning the momentum and the kinetic energy of the ISS is correct?

  1. Both the momentum and the kinetic energy vary along the orbital path.
  2. Both the momentum and the kinetic energy are constant along the orbital path.
  3. The momentum is constant, but the kinetic energy changes throughout the orbital path.
  4. The momentum changes, but the kinetic energy remains constant throughout the orbital path.
Show Answers Only

\(D\)

Show Worked Solution

→ The magnitude of both the kinetic energy and momentum of the ISS remains constant.

→ However, momentum is a vector and as the direction of the ISS is continually changing so is the direction of the momentum of the ISS.

\(\Rightarrow D\)

♦♦ Mean mark 31%.
COMMENT: Students need to ensure they consider the direction of vector quantities as well as the magnitudes.

PHYSICS, M8 2021 VCE 19

A simplified diagram of some of the energy levels of an atom is shown in Figure 16.
 

  1. Identify the transition on the energy level diagram in Figure 16 that would result in the emission of a 565 nm photon. Show your working.   (2 marks)

  1. A sample of the atoms is excited into the 9.8 eV state and a line spectrum is observed as the states decay. Assume that all possible transitions occur.

    What is the total number of lines in the spectrum? Explain your answer. You may use the diagram below to support your answer.   (2 marks)
     

Show Answers Only

a.    The energy transition from 8.9 eV to 6.7 eV.

b.    9 spectral lines.

Show Worked Solution

a.    \(E=hf=\dfrac{hc}{\lambda}=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{565 \times 10^{-9}}=3.52 \times 10^{-19}\ \text{J}\)

\(\text{Convert to eV}\ =\dfrac{3.52 \times 10^{-19}}{1.602 \times 10^{-19}}=2.2\ \text{eV}\)

→ The energy transition from 8.9 eV to 6.7 eV will result in the emission of a 565 nm photon.

→ Note: This could have also been shown on the diagram as a downwards arrow from 8.9 eV to 6.7 eV.
 

♦ Mean mark (a) 50%.

b.    Each photon emitted that has its own unique energy will result in a spectrum line.

→ There are 4 possible transitions from 9.8 eV, 3 from 8.9 eV, 2 from 6.7 eV and 1 from 4.9 eV, resulting in 10 different transitions.

→ However, the transition from 9.8 eV to 4.9 eV and 4.9 eV to 0 eV will produce a photon of the same energy, hence they will have the same spectral line. 

→ Therefore, there will be 9 lines in the spectrum.

♦♦ Mean mark (b) 31%.

Calculus, 2ADV C4 2023 MET2 6 MC

Suppose that \(\displaystyle \int_{3}^{10} f(x)\,dx=C\)  and  \(\displaystyle \int_{7}^{10} f(x)\,dx=D\). The value of \(\displaystyle \int_{7}^{3} f(x)\,dx\) is

  1. \(C+D\)
  2. \(C+D-3\)
  3. \(C-D\)
  4. \(D-C\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Given }\displaystyle \int_{3}^{10} f(x)\,dx=C\ \ \text{and}\ \displaystyle \int_{7}^{10} f(x)\,dx=D\)

\(\text{We can deduce:}\)

 \(\displaystyle \int_{3}^{10} f(x)\,dx\) \(=\displaystyle \int_{3}^{7} f(x)\,dx+\displaystyle \int_{7}^{10} f(x)\,dx\)
\(C\) \(=\displaystyle \int_{3}^{7} f(x)\,dx+D\)
\(C-D\) \(=\displaystyle \int_{3}^{7} f(x)\,dx\)
\(\therefore\ \displaystyle \int_{7}^{3} f(x)\,dx\) \(=D-C\)

 
\(\Rightarrow D\)


♦ Mean mark 49%.
MARKER’S COMMENT: 41% of students chose option C incorrectly assuming \(\int_{7}^3 f(x)\,dx=\int_{3}^7 f(x)\,dx\).

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

PHYSICS, M7 2021 VCE 16

Light can be described by a wave model and also by a particle (or photon) model. The rapid emission of photoelectrons at very low light intensities supports one of these models but not the other.

Identify the model that is supported, giving a reason for your answer.

Show Answers Only

→ Model supported: Particle (photon) model

Reasons could include one of the following:

→ Each electron ejection corresponds to the absorption of a single photon.

→ The immediate ejection of a photoelectron corresponds to its direct interaction with the initial photon.

→ Contrary to the wave theory, which suggests that energy accumulates gradually, the emission of photoelectrons at very low light intensities demonstrates that energy is delivered in discrete quanta.

Show Worked Solution

→ Model supported: Particle (photon) model

Reasons could include one of the following:

→ Each electron ejection corresponds to the absorption of a single photon.

→ The immediate ejection of a photoelectron corresponds to its direct interaction with the initial photon.

→ Contrary to the wave theory, which suggests that energy accumulates gradually, the emission of photoelectrons at very low light intensities demonstrates that energy is delivered in discrete quanta.

♦ Mean mark 42%.

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

  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^(”)(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 19 MC

A box is formed from a rectangular sheet of cardboard, which has a width of `a` units and a length of `b` units, by first cutting out squares of side length `x` units from each corner and then folding upwards to form a container with an open top.

The maximum volume of the box occurs when `x` is equal to

  1. `\frac{a-b+\sqrt{a^2-a b+b^2}}{6}`
  2. `\frac{a+b+\sqrt{a^2-a b+b^2}}{6}`
  3. `\frac{a-b-\sqrt{a^2-a b+b^2}}{6}`
  4. `\frac{a+b-\sqrt{a^2-a b+b^2}}{6}`
  5. `\frac{a+b-\sqrt{a^2-2 a b+b^2}}{6}`
Show Answers Only

`D`

Show Worked Solution

`V(x)=x(b-2 x)(a-2 x)=abx-2(a+b)x^2+4x^3`

Maximum occurs when `V\^{\prime} (x)=0`

`V\^{\prime} (x)=ab-4(a+b)x+12x^2=0`

Solving for `x` using CAS.

`x=\frac{a+b \+ \sqrt{a^2-a b+b^2}}{6}`  or  `x=\frac{a+b \- \sqrt{a^2-a b+b^2}}{6}`

Testing shape of the cubic using `b = 2` and `a = 1`

Maximum occurs at the smaller value of `x`

`:.\  x=\frac{a+b \- \sqrt{a^2-a b+b^2}}{6}`

`=>D`


♦♦ Mean mark 34%.
MARKER’S COMMENT: 34% of students incorrectly chose option B. 

Probability, MET2 2022 VCAA 18 MC

If `X` is a binomial random variable where `n=20, p=0.88` and `\text{Pr}(X \geq 16|X\geq a)=0.9175`, correct to four decimal places, then `a` is equal to

  1. 11
  2. 12
  3. 13
  4. 14
  5. 15
Show Answers Only

`B`

Show Worked Solution

`X \~ \text{Bi}(20,0.88)`

`\text{Pr}(X \geq 16|X\geq a)=0.9175`

`\frac{\text{Pr}(X \geq 16 \cap X \geq a)}{\text{Pr}(X \geq a)} \approx 0.9175`
 

All options are `< 16`

`:.\ \frac{\text{Pr}(X \geq 16)}{\text{Pr}(X \geq a)} \approx 0.9175`
 

Testing options using CAS

Option A: `\ \frac{\text{Pr}(X \geq 16)}{\text{Pr}(X \geq 11)} \approx 0.9173`

Option B: `\ \frac{\text{Pr}(X \geq 16)}{\text{Pr}(X \geq 12)} \approx 0.9175`

Option C: `\ \frac{\text{Pr}(X \geq 16)}{\text{Pr}(X \geq 13)} \approx 0.9186`

Option D: `\ \frac{\text{Pr}(X \geq 16)}{\text{Pr}(X \geq 14)} \approx 0.9235`

Option E: `\ \frac{\text{Pr}(X \geq 16)}{\text{Pr}(X \geq 15)} \approx 0.9418`
  

`=>B`


♦♦ Mean mark 47%.

Calculus, MET2 2022 VCAA 17 MC

A function `g` is continuous on the domain `x \in[a, b]` and has the following properties:

  • The average rate of change of `g` between `x=a` and `x=b` is positive.
  • The instantaneous rate of change of `g` at `x=\frac{a+b}{2}` is negative.

Therefore, on the interval `x \in[a, b]`, the function must be

  1. many-to-one.
  2. one-to-many.
  3. one-to-one.
  4. strictly decreasing.
  5. strictly increasing.
Show Answers Only

`A`

Show Worked Solution

The first property shows that the point (`a` , __ ) must be below (`b` , __ ) to give a positive rate of change.

The second property gives the midpoint of `a` and `b`. But this point has a negative instantaneous rate of change.

Therefore, this shows that the function `g(x)` has turning points and is in the shape of a cubic, so many-to-one is the correct description.

 
`=>A`


♦♦ Mean mark 39%.

Graphs, MET2 2022 VCAA 13 MC

The function `f(x)=\log _e\left(\frac{x+a}{x-a}\right)`, where `a` is a positive real constant, has the maximal domain

  1. `[-a, a]`
  2. `(-a, a)`
  3. `R \backslash[-a, a]`
  4. `R \backslash(-a, a)`
  5. `R`
Show Answers Only

`C`

Show Worked Solution

`f(x)=\log _e\left(\frac{x+a}{x-a}\right),\ a>0`

For `\frac{x+a}{x-a}>0` either both `x + a` and `x – a` are positive and `x > a` or both are negative and `x < -a`.

`:.` maximal domain occurs when `\ x \in R \backslash[-a, a]`
 

`=>C`


♦♦ Mean mark 39%.

Probability, MET2 2022 VCAA 12 MC

A bag contains three red pens and `x` black pens. Two pens are randomly drawn from the bag without replacement. The probability of drawing a pen of each colour is equal to

  1. `\frac{6 x}{(2+x)(3+x)}`
  2. `\frac{3 x}{(2+x)(3+x)}`
  3. `\frac{x}{2+x}`
  4. `\frac{3+x}{(2+x)(3+x)}`
  5. `\frac{3+x}{5+2 x}`
Show Answers Only

`A`

Show Worked Solution
Pr`(RB)` + Pr`(BR)` `= \frac{3}{3+x} \times \frac{x}{2+x}+\frac{x}{3+x} \times \frac{3}{2+x}`  
  `= \frac{6 x}{(2+x)(3+x)}`  

 
`=>A`


♦ Mean mark 52%.

 

Graphs, MET2 2022 VCAA 9 MC

Let `f:[0, \infty) \rightarrow R, f(x)=\sqrt{2 x+1}`.

The shortest distance, `d`, from the origin to the point `(x, y)` on the graph of `f` is given by

  1. `d=x^2+2 x+1`
  2. `d=x^2+\sqrt{2 x+1}`
  3. `d=\sqrt{x^2-2 x+1}`
  4. `d=x+1`
  5. `d=2 x+1`
Show Answers Only

`D`

Show Worked Solution

Straight line distance is the shortest.

`:.` Therefore, using pythagoras and given `f(x) = \sqrt{2x+1}` 

`d^2` `= x^2 + y^2`  
  `=x^2 + (sqrt{2x+1})^2`  
  `= x^2 + 2x +1`  
  `=(x + 1)^2`  
`d` `=sqrt((x + 1)^2)`  
  `= x+1`  

  
`=>D`


♦ Mean mark 50%.

Functions, MET2 2022 VCAA 6 MC

Which of the pairs of functions below are not inverse functions?

  1. \( \begin {cases}f(x)=5x+3         &\ \ x\in R \\ g(x)=\dfrac{x-3}{5}         &\ \ x \in R \\ \end{cases}\)
     
  2. \( \begin {cases}f(x)=\frac{2}{3}x+2         &\ \ x\in R \\ g(x)=\frac{3}{2}x-3         &\ \ x \in R \\ \end{cases}\)
     
  3. \( \begin {cases}f(x)=x^2         &\ \ x<0 \\ g(x)=\sqrt{x}         &\ \ x >0 \\ \end{cases}\)
     
  4. \( \begin {cases}f(x)=\dfrac{1}{x}        &\ \ x\neq 0 \\ g(x)=\dfrac{1}{x}         &\ \ x \neq 0 \\ \end{cases}\)
     
  5. \( \begin {cases}f(x)=\log_e(x)+1        &\ \ x>0 \\ g(x)=e^{x-1}        &\ \ x \in R \\ \end{cases}\)
Show Answers Only

\(C\)

Show Worked Solution

\(\text{Graphically, it can be seen that } f(x)=x^2\ \text{and }g(x)=\sqrt{x}\ \text{are not}\)

\(\text{inverse functions as }g(x)\ \text{is not a reflection of }f(x)\ \text{in the line }y=x.\)

\(g(x)\ \text{should be the curve drawn as } h(x)=-\sqrt{x}\ \text{below}.\)

\(\Rightarrow C\)


♦ Mean mark 47%.

PHYSICS, M7 2021 VCE 13

In Young's double-slit experiment, the distance between two slits, S\(_1\) and S\(_2\), is 2.0 mm. The slits are 1.0 m from a screen on which an interference pattern is observed, as shown in Figure 13a. Figure 13b shows the central maximum of the observed interference pattern.
 

  1. If a laser with a wavelength of 620 nm is used to illuminate the two slits, what would be the distance between two successive dark bands? Show your working.  (2 marks)
  1. Explain how this experiment supports the wave model of light.  (2 marks)

Show Answers Only

a.    \(3.1 \times 10^{-4}\ \text{m}\)

b.    The experiment supports the wave theory as follows:

→ Young’s double slit experiment demonstrates how light will refract and form an interference pattern against a screen.

→ Since interference is a wave phenomenon, the experiment supports the wave model of light.

Show Worked Solution

a.    \(d\,\sin \theta=m \lambda\ \ \text{and}\ \ \sin \theta=\dfrac{\Delta x}{D}\)

→ \(\Delta x\) is the distance between two successive dark bands and \(D\) is the distance from the slits to the screen.

\(\dfrac{d\Delta x}{D}\) \(=m \lambda\)  
\(\Delta x\) \(=\dfrac{D\lambda}{d}\)  
  \(=\dfrac{1 \times 620 \times 10^{-9}}{2 \times 10^{-3}}\)  
  \(=3.1 \times 10^{-4}\ \text{m}\)  
♦ Mean mark (a) 51%.

 
b.    The experiment supports the wave theory as follows:

→ Young’s double slit experiment demonstrates how light will refract and form an interference pattern against a screen.

→ Since interference is a wave phenomenon, the experiment supports the wave model of light.

PHYSICS, M6 2021 VCE 6

Figure 6 shows a simple AC generator. A mechanical energy source rotates the loop smoothly at 50 revolutions per second and the loop generates a voltage of 6 V. The magnetic field, \(B\), is constant and uniform. The direction of rotation is as shown in Figure 6.
 

  1. Sketch the output EMF between \(\text{P}\) and \(\text{Q}\) versus time, \(t\), on the grid below, starting with the loop in the position shown in Figure 6. Show at least two complete revolutions, and include the voltage on the vertical axis and a time scale on the horizontal axis.   (3 marks)
     

  1. Describe the function of the slip rings shown in Figure 6.   (1 mark)

  1.  i. How could the AC generator shown in Figure 6 be changed to a DC generator?   (1 mark)

 ii.  Sketch the output EMF versus time, \(t\), for this DC generator for at least two complete revolutions on the grid below. Include the voltage on the vertical axis and a time scale on the horizontal axis. (2 marks)
 

Show Answers Only

a.   

         

b.    The function of the slip rings:

→ Produce an AC voltage output by maintaining a constant connection between the loop and the external circuit.
 

c.i.   Changing the slip rings to a split ring commutator.

c.ii. 

         

Show Worked Solution

a.    \(\text{Period}\ (T)=\dfrac{1}{f}=\dfrac{1}{50}=0.02\ \text{s}\).

b.    The function of the slip rings:

→ Produce an AC voltage output by maintaining a constant connection between the loop and the external circuit.

♦ Mean mark (b) 54%.

c.i.   Changing the slip rings to a split ring commutator.
 

c.ii.  The output of a DC generator is positive only.

   

♦ Mean mark (c)(i) 55%.

PHYSICS, M6 2021 VCE 7

The generator of an electrical power plant delivers 500 MW to external transmission lines when operating at 25 kV. The generator's voltage is stepped up to 500 kV for transmission and stepped down to 240 V 100 km away (for domestic use). The overhead transmission lines have a total resistance of 30.0 \(\Omega\). Assume that all transformers are ideal.

  1. Explain why the voltage is stepped up for transmission along the overhead transmission lines.   (2 marks)
  1. Calculate the current in the overhead transmission lines. Show your working.   (2 marks)
  1. Determine the maximum power available for domestic use at 240 V. Show all your working.   (3 marks)

Show Answers Only

a.    → \(P=IV\)

→ The voltage is stepped up for transmission in overhead transmission lines to reduce the current while maintaining a constant power output.

→ Since  \(P_{\text{loss}}=I^2R\), if the current is reduced, the power loss will is reduced significantly.

b.    \(I=1000\ \text{A}\)

c.    \(P_{\text{avail}}=470\ \text{MW}\)

Show Worked Solution

a.    → \(P=IV\)

→ The voltage is stepped up for transmission in overhead transmission lines to reduce the current while maintaining a constant power output.

→ Since  \(P_{\text{loss}}=I^2R\), if the current is reduced, the power loss will is reduced significantly.
 

♦♦ Mean mark (a) 30%.
COMMENT: Students must state that the power is constant when stepping up the voltage to decrease the current.

b.    → The power in the lines is 500 MW.

→ Voltage through the transmission lines is 500 kV.

→ Using \(P=IV:\)

\(I=\dfrac{P}{V}=\dfrac{500 \times 10^6}{500 \times 10^3}=1000\ \text{A}\)
 

♦ Mean mark (b) 43%.

c.   → \(P_{\text{loss}}=I^2R=(1000)^2 \times 30=30\ \text{MW}\)

→ \(\text{Power (delivered)}\ =500\ \text{MW}-30\ \text{MW}=470\ \text{MW}\)

♦♦ Mean mark (c) 38%.

Functions, 2ADV F2 2022 SPEC2 3 MC

The graph of  `y=\frac{x^2+2x+c}{x^2-4}`, where `c \in R`, will always have

  1. two vertical asymptotes and one horizontal asymptote.
  2. a vertical asymptote with equation `x=-2` and one horizontal asymptote with equation `y=1`.
  3. one horizontal asymptote with equation `y=1` and only one vertical asymptote with equation `x=2`.
  4. a horizontal asymptote with equation `y=1` and at least one vertical asymptote.
Show Answers Only

`D`

Show Worked Solution

`y=\frac{x^2+2x+c}{x^2-4}\ \ =>\ \ y=\frac{x^2+2x+c}{(x-2)(x+2)}`

`text{Vertical asymptotes:}\ x=2, \ x=-2`

`->\ text{However, if}\ c=0,\ \text{only 1 vertical asymptote at}\ \ x=2.`

`text{Horizontal asymptote:}\ y=1`

`=>D`


♦♦ Mean mark 38%.
41% of students chose `A` and did not consider when `c = 0`

Graphs, SPEC2 2022 VCAA 3 MC

The graph of  `y=\frac{x^2+2x+c}{x^2-4}`, where `c \in R`, will always have

  1. two vertical asymptotes and one horizontal asymptote.
  2. two horizontal asymptotes and one vertical asymptote.
  3. a vertical asymptote with equation `x=-2` and one horizontal asymptote with equation `y=1`.
  4. one horizontal asymptote with equation `y=1` and only one vertical asymptote with equation `x=2`.
  5. a horizontal asymptote with equation `y=1` and at least one vertical asymptote.
Show Answers Only

`E`

Show Worked Solution

`y=\frac{x^2+2x+c}{x^2-4}\ \ =>\ \ y=\frac{x^2+2x+c}{(x-2)(x+2)}`

`text{Vertical asymptotes:}\ x=2, \ x=-2`

`->\ text{However, if}\ c=0,\ \text{only 1 vertical asymptote at}\ \ x=2.`

`text{Horizontal asymptote:}\ y=1`

`text{Using CAS to graph for different values of}\ c:`

`=>E`


♦♦ Mean mark 38%.
41% of students chose `A` and did not consider when `c = 0`

Calculus, MET2 2023 VCAA 3

Consider the function \(g:R \to R, g(x)=2^x+5\).

  1. State the value of \(\lim\limits_{x\to -\infty} g(x)\).   (1 mark)
  2. The derivative, \(g'(x)\), can be expressed in the form \(g'(x)=k\times 2^x\).
  3. Find the real number \(k\).   (1 mark)
    1. Let \(a\) be a real number. Find, in terms of \(a\), the equation of the tangent to \(g\) at the point \(\big(a, g(a)\big)\).   (1 mark)
    1. Hence, or otherwise, find the equation of the tangent to \(g\) that passes through the origin, correct to three decimal places.   (2 marks)

Let \(h:R\to R, h(x)=2^x-x^2\).

  1. Find the coordinates of the point of inflection for \(h\), correct to two decimal places.   (1 mark)
  2. Find the largest interval of \(x\) values for which \(h\) is strictly decreasing.
  3. Give your answer correct to two decimal places.   (1 mark)
  4. Apply Newton's method, with an initial estimate of \(x_0=0\), to find an approximate \(x\)-intercept of \(h\).
  5. Write the estimates \(x_1, x_2,\) and \(x_3\) in the table below, correct to three decimal places.   (2 marks)
      

    \begin{array} {|c|c|}
    \hline
    \rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
    \hline
    \rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \end{array}
  6. For the function \(h\), explain why a solution to the equation \(\log_e(2)\times (2^x)-2x=0\) should not be used as an initial estimate \(x_0\) in Newton's method.   (1 mark)
  7. There is a positive real number \(n\) for which the function \(f(x)=n^x-x^n\) has a local minimum on the \(x\)-axis.
  8. Find this value of \(n\).   (2 marks)
Show Answers Only

a.    \(5\)

b.    \(\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)

ci.  \(y=2^a\ \log_{e}{(2)x}-(a\ \log_{e}{(2)}-1)\times2^a+5\)

\(\text{or}\ \ y=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

cii. \(y=4.255x\)

d. \((2.06 , -0.07)\)

e. \([0.49, 3.21]\)

f. 

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  -1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.773 \\
\hline
\end{array}

g. \(\text{See worked solution.}\)

h. \(n=e\)

Show Worked Solution

a.    \(\text{As }x\to -\infty,\ \ 2^x\to 0\)

\(\therefore\ 2^x+5\to 5\)
  

b.     \(g(x)\) \(=2^x+5\)
    \(=\Big(e^{\log_{e}{2}}\Big)^x\)
    \(=e\ ^{x\log_{e}{2}}+5\)
  \(g'(x)\) \(=\log_{e}{2}\times e\ ^{x\log_{e}{2}}\)
     \(=\log_{e}{2}\times 2^x\)
  \(\therefore\ k\) \(=\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)
   
ci.  \(\text{Tangent at}\ (a, g(a)):\)

\(y-(2^a+5)\) \(=\log_{e}{2}\times2^a(x-a)\)
\(\therefore\ y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

♦♦ Mean mark (c)(i) 50%.

cii.  \(\text{Substitute }(0, 0)\ \text{into equation from c(i) to find}\ a\)

\( y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=2^a\ \log_{e}{(2)\times 0}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

  
\(\text{Solve for }a\text{ using CAS }\rightarrow\ a\approx 2.61784\dots\)

\(\text{Equation of tangent when }\ a\approx 2.6178\)

\( y\) \(=2^{2.6178..}\ \log_{e}{(2)x}+0\)
\(\therefore\  y\) \(=4.255x\)

♦♦♦ Mean mark (c)(ii) 30%.
MARKER’S COMMENT: Some students did not substitute (0, 0) into the correct equation or did not find the value of \(a\) and used \(a=0\).
d.     \(h(x)\) \(=2^x-x^2\)
  \(h'(x)\) \(=\log_{e}{(2)}\cdot 2^x-2x\ \ \text{(Using CAS)}\)
  \(h”(x)\) \(=(\log_{e}{(2)})^2\cdot 2^x-2\ \ \text{(Using CAS)}\)

\(\text{Solving }h”(x)=0\ \text{using CAS }\rightarrow\ x\approx 2.05753\dots\)

\(\text{Substituting into }h(x)\ \rightarrow\ h(2.05753\dots)\approx-0.070703\dots\)

\(\therefore\ \text{Point of inflection at }(2.06 , -0.07)\ \text{ correct to 2 decimal places.}\)

e.    \(\text{From graph (CAS), }h(x)\ \text{is strictly decreasing between the 2 turning points.}\)

\(\therefore\ \text{Largest interval includes endpoints and is given by }\rightarrow\ [0.49, 3.21]\)


♦♦ Mean mark (e) 40%.
MARKER’S COMMENT: Round brackets were often used which were incorrect as endpoints were included. Some responses showed the interval where the function was strictly increasing.

f.    \(\text{Newton’s Method }\Rightarrow\  x_a-\dfrac{h(x_a)}{h'(x_a)}\) 

\(\text{for }a=0, 1, 2, 3\ \text{given an initial estimation for }x_0=0\)

\(h(x)=2x-x^2\ \text{and }h'(x)=\ln{2}\times 2^x-2x\)

\begin{array} {|c|l|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  0-\dfrac{2^0-2\times 0}{\ln2\times 2^0\times 0}=-1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -1.433-\dfrac{2^{-1.433}-2\times -1.433}{\ln2\times 2^{-1.433}\times -1.433}=-0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.897-\dfrac{2^{-0.897}-2\times -0.897}{\ln2\times 2^{-0.897}\times -0.897}=-0.773 \\
\hline
\end{array}

g.    \(\text{The denominator in Newton’s Method is}\ h'(x)=\log_{e}{(2)}\cdot 2^x-2x\)

\(\text{and the calculation will be undefined if }h'(x)=0\ \text{as the tangent lines are horizontal}.\)

\(\therefore\ \text{The solution to }h'(x)=0\ \text{cannot be used for }x_0.\)

♦♦♦ Mean mark (g) 20%.

h.    \(\text{For a local minimum }f(x)=0\)

\(\rightarrow\ n^x-x^n=0\)

\(\rightarrow\ n^x=x^n\ \ \ (1)\)

\(\text{Also for a local minimum }f'(x)=0\)

\(\rightarrow\ \ln(n)\cdot n^x-nx^{n-1}=0\ \ \ (2)\)

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

\(\ln(n)\cdot x^n-nx^{n-1}=0\) 

\(x^n\Big(\ln(n)-\dfrac{n}{x}\Big)=0\)

\(\therefore\ x^n=0\ \text{or }\ \ln(n)=\dfrac{n}{x}\)

\(x=0\ \text{or }x=\dfrac{n}{\ln(n)}\)

\(\therefore\ n=e\)


♦♦♦ Mean mark (h) 10%.
MARKER’S COMMENT: Many students showed that \(f'(x)=0\) but failed to couple it with \(f(x)=0\). Ensure exact values are given where indicated not approximations.

Calculus, EXT1 C3 2022 SPEC1 10

Let `f(x)=\sec (4 x)`.

  1. Sketch the graph of `f` for `x \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]` on the set of axes below. Label any asymptotes with their equations and label any turning points and the endpoints with their coordinates.   (3 marks)
      

      
  2. The graph of  `y=f(x)` for `x \in\left[-\frac{\pi}{24}, \frac{\pi}{48}\right]` is rotated about the `x`-axis to form a solid of revolution.
    Find the volume of this solid. Give your answer in the form `\frac{(a-\sqrt{b}) \pi}{c}`, where `a`, `b`, `c in R`.   (3 marks)
Show Answers Only

a.  
       

b.   `\frac{(3-\sqrt{3}) \pi}{6}`

Show Worked Solution

a.      
       


♦ Mean mark (a) 49%.
b.    `V` `=\pi \int_{-\frac{\pi}{24}}^{\frac{\pi}{48}} \sec ^2(4 x)\ dx`
    `=\frac{\pi}{4}[\tan (4 x)]_{-\frac{\pi}{24}}^{\frac{\pi}{48}}`
    `=\frac{\pi}{4} \tan \left(\frac{\pi}{12}\right)-\frac{\pi}{4} \tan \left(-\frac{\pi}{6}\right)`
    `=\frac{\pi}{4} \tan \left(\frac{\pi}{3}-\frac{\pi}{4}\right)-\frac{\pi}{4} \xx -\frac{1}{\sqrt{3}}`
    `=\frac{\pi}{4} \left(\frac{sqrt3-1}{1+sqrt3} xx \frac{1-sqrt3}{1-sqrt3}\right)+\frac{\pi}{4sqrt3}`
    `=\frac{\pi}{4} \left(\frac{sqrt3-3-1+sqrt3}{-2}\right)+\frac{\pi}{4sqrt3}`
    `=\frac{\pi}{4}(2-\sqrt{3}) +\frac{\pi}{4sqrt3}`
    `=\frac{\pi(2 \sqrt{3}-3+1)}{4 \sqrt{3}}`
    `=\frac{(6-2 \sqrt{3}) \pi}{12}`
    `=\frac{(3-\sqrt{3}) \pi}{6}`

♦ Mean mark (b) 55%.

Calculus, SPEC1 2022 VCAA 10

Let `f(x)=\sec (4 x)`.

  1. Sketch the graph of `f` for `x \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]` on the set of axes below. Label any asymptotes with their equations and label any turning points and the endpoints with their coordinates.   (3 marks)
      

      
  2. The graph of  `y=f(x)` for `x \in\left[-\frac{\pi}{24}, \frac{\pi}{48}\right]` is rotated about the `x`-axis to form a solid of revolution.
    Find the volume of this solid. Give your answer in the form `\frac{(a-\sqrt{b}) \pi}{c}`, where `a`, `b`, `c in R`.   (3 marks)
Show Answers Only

a.  
       

b.   `\frac{(3-\sqrt{3}) \pi}{6}`

Show Worked Solution

a.      
       


♦ Mean mark (a) 49%.
b.    `V` `=\pi \int_{-\frac{\pi}{24}}^{\frac{\pi}{48}} \sec ^2(4 x)\ dx`
    `=\frac{\pi}{4}[\tan (4 x)]_{-\frac{\pi}{24}}^{\frac{\pi}{48}}`
    `=\frac{\pi}{4} \tan \left(\frac{\pi}{12}\right)-\frac{\pi}{4} \tan \left(-\frac{\pi}{6}\right)`
    `=\frac{\pi}{4} \tan \left(\frac{\pi}{3}-\frac{\pi}{4}\right)-\frac{\pi}{4} \xx -\frac{1}{\sqrt{3}}`
    `=\frac{\pi}{4} \tan \left(\frac{sqrt3-1}{1+sqrt3} xx \frac{1-sqrt3}{1-sqrt3}\right)+\frac{\pi}{4sqrt3}`
    `=\frac{\pi}{4} \tan \left(\frac{sqrt3-3-1+sqrt3}{-2}\right)+\frac{\pi}{4sqrt3}`
    `=\frac{\pi}{4}(2-\sqrt{3}) +\frac{\pi}{4sqrt3}`
    `=\frac{\pi(2 \sqrt{3}-3+1)}{4 \sqrt{3}}`
    `=\frac{(6-2 \sqrt{3}) \pi}{12}`
    `=\frac{(3-\sqrt{3}) \pi}{6}`

♦ Mean mark (b) 55%.

Calculus, EXT2 C1 2022 SPEC1 9

Given that  `f^{\prime}(x)=\frac{\cos (2 x)}{\sin ^3(2 x)}`  and  `f((pi)/(8))=(3)/(4)`, find `f(x)`.   (4 marks)

Show Answers Only

`f(x)=\frac{-1}{4 \sin ^2(2 x)}+\frac{5}{4}=-\frac{1}{4} \text{cosec}^2(2 x)+\frac{5}{4}`

Show Worked Solution

`text{Let}\ \ u=sin(2x)\ \ =>\ \ {du}/{dx}=2cos(2x)`

`\int \frac{\cos (2 x)}{\sin ^3(2 x)} d x` `=\frac{1}{2} \int \frac{d u}{u^3}`  
  `= -\frac{1}{4} u^{-2}+c`  
  `=-\frac{1}{4} \ xx \frac{1}{\sin ^2(2 x)}+c`  

 
`text{When}\ \ x = pi/8:`

`-\frac{1}{4} \cdot \frac{1}{(\frac{1}{sqrt{2}})^2}+c` `=\frac{3}{4}`  
`-\frac{1}{4} \cdot 2+c` `=\frac{3}{4}`  
`\Rightarrow c` `=\frac{3}{4}+\frac{2}{4}=5/4`  

 
`:. \  f(x)=\frac{-1}{4 \sin ^2(2 x)}+\frac{5}{4}=-\frac{1}{4} \text{cosec}^2(2 x)+\frac{5}{4}`


♦ Mean mark 50%.

Calculus, SPEC1 2022 VCAA 9

Given that  `f^{\prime}(x)=\frac{\cos (2 x)}{\sin ^3(2 x)}`  and  `f((pi)/(8))=(3)/(4)`, find `f(x)`.   (4 marks)

Show Answers Only

`f(x)=\frac{-1}{4 \sin ^2(2 x)}+\frac{5}{4}=-\frac{1}{4} \text{cosec}^2(2 x)+\frac{5}{4}`

Show Worked Solution

`text{Let}\ \ u=sin(2x)\ \ =>\ \ {du}/{dx}=2cos(2x)`

`\int \frac{\cos (2 x)}{\sin ^3(2 x)} d x` `=\frac{1}{2} \int \frac{d u}{u^3}`  
  `= -\frac{1}{4} u^{-2}+c`  
  `=-\frac{1}{4} \ xx \frac{1}{\sin ^2(2 x)}+c`  

 
`text{When}\ \ x = pi/8:`

`-\frac{1}{4} \cdot \frac{1}{(\frac{1}{sqrt{2}})^2}+c` `=\frac{3}{4}`  
`-\frac{1}{4} \cdot 2+c` `=\frac{3}{4}`  
`\Rightarrow c` `=\frac{3}{4}+\frac{2}{4}=5/4`  

 
`:. \  f(x)=\frac{-1}{4 \sin ^2(2 x)}+\frac{5}{4}=-\frac{1}{4} \text{cosec}^2(2 x)+\frac{5}{4}`


♦ Mean mark 50%.

Calculus, SPEC1 2022 VCAA 7

A curve has equation `x cos(x+y)=(pi)/(48)`.

Find the gradient of the curve at the point `((pi)/(24),(7pi)/(24))`. Give your answer in the form `(asqrtb-pi)/(pi)`, where `a,b in Z`.   (3 marks)

Show Answers Only

`(8sqrt3-pi)/pi`

Show Worked Solution

`text{Using implicit differentiation and the chain and product rules:}`

`d/(dx)(cos(x + y))`

`text{Let}\ \ u = x + y\ \ \=>\ \ y = cos(u)` 

`(du)/(dx) = 1 +dy/dx,\ \ (dy)/(du) = -sin(u)`

`cos (x+y)-x\ sin (x+y)(dy/dx+1)` `=0`  
`cos (x+y)-x\ sin(x+y)(dy/dx)-x\ sin(x+y)` `=0`  
`x\ sin(x+y)(dy/dx)` `=cos (x+y)-x\ sin(x+y)`  
`dy/dx` `=(cos (x+y))/(x\ sin(x+y)) – (x\ sin(x+y))/(x\ sin(x+y))`  
  `=(cos (x+y))/(x\ sin(x+y))-1`  

 
`text{At}\ ((pi)/(24),(7pi)/(24))\ \ \=>\ \ x+y = (pi)/(24)+(7pi)/(24) = (pi)/(3)`

`dy/dx` `=cos(pi/3)÷[(pi)/(24)sin(pi/3)]-1`  
  `=(1/2)/((pi)/(24) xx sqrt3/2)-1`  
  `=24/(sqrt3pi)-1`  
  `=(24-sqrt3pi)/(sqrt3pi) xx (sqrt3/sqrt3)`  
  `=(24sqrt3-3pi)/(3pi)`  
  `=(8sqrt3-pi)/pi`  

♦ Mean mark 50%.

Statistics, SPEC1 2022 VCAA 3

The time taken by a coffee machine to dispense a cup of coffee varies normally with a mean of 10 seconds and a standard deviation of 1.5 seconds.

Find the probability that more than 34 seconds is needed to dispense a total of four cups of coffee. Give your answer correct to two decimal places.   (2 marks) 

Show Answers Only

`0.98`

Show Worked Solution

`\text{1 cup:}\ \ mu=10,\ \ sigma=1.5, \ \text{Var(X)}\ =1.5^2 = 2.25`

`\text{4 cups:}\ \ mu=4 xx 10= 40,\ \text{Var(X)}\ = 4 xx 2.25=9, \ sigma = \sqrt{9}=3`

`Z\ ~\ N(0,1)`

`text{Pr}(Z>(34-40)/(3))` `= text{Pr}(Z> -2)`  
  `~~ 0.975`  
  `=0.98\ \text{(2 d.p.)}`  

♦♦ Mean mark 36%.

Vectors, SPEC2 2023 VCAA 5

The points with coordinates \(A(1,1,2), B(1,2,3)\) and \(C(3,2,4)\) all lie in a plane \(\Pi\).

  1. Find the vectors \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\), and hence show that the area of triangle \(A B C\) is 1.5 square units.  (2 marks)

  2. Find the shortest distance from point \(B\) to the line segment \(A C\).  (2 marks)

A second plane, \(\psi\), has the Cartesian equation  \(2 x-2 y-z=-18\).

  1. At what acute angle does the line given by \(\underset{\sim}{ r }(t)=3 \underset{\sim}{ i }+2 \underset{\sim}{ j }+4 \underset{\sim}{ k }+t(\underset{\sim}{ i }-2 \underset{\sim}{ j }+2\underset{\sim}{ k }), t \in R\), intersect the plane \(\psi\) ? Give your answer in degrees correct to the nearest degree.  (2 marks)

A line \(L\) passes through the origin and is normal to the plane \(\psi\). The line \(L\) intersects \(\psi\) at a point \(D\).

  1. Write down an equation of the line \(L\) in parametric form.  (1 mark)

  2. Find the shortest distance from the origin to the plane \(\psi\).  (2 marks)

  3. Find the coordinates of point \(D\).  (2 marks)

Show Answers Only

a.     \(\overrightarrow{A B}  =\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=\left(\begin{array}{l}0 \\ 1 \\ 1\end{array}\right)\)
  \(\overrightarrow{A C}  =\overrightarrow{O C}-\overrightarrow{O A}=\left(\begin{array}{l}3 \\ 2 \\ 4\end{array}\right)-\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 1 \\ 2\end{array}\right)\)

\(\text{Area}=\dfrac{1}{2}\ \bigg|\overrightarrow{AB} \times \overrightarrow{AC}\bigg|=\dfrac{1}{2}\ \left|\begin{array}{ccc}\underset{\sim}{i} & \underset{\sim}{j} & \underset{\sim}{k} \\ 0 & 1 & 1 \\ 2 & 1 & 2\end{array}\right|=\dfrac{1}{2}\ \bigg|\underset{\sim}{i}+2 \underset{\sim}{j}-2 \underset{\sim}{k}\bigg|=\dfrac{3}{2}\)

b.    \(\text {Shortest distance }=1 \text { unit }\)

c.     \(26^{\circ}\)

d.    \(\underset{\sim}{r}(t)=\underset{\sim}{n} \times t=2 t \underset{\sim}{i}-2 t \underset{\sim}{j}-t \underset{\sim}{k}\)

\(x=2 t, y=-2 t, z=-t \quad \text{(also accepted)}\)

e.    \(\text {Shortest distance }=6\)

f.    \(D(-4,4,2)\)

Show Worked Solution

a.     \(\overrightarrow{A B}  =\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=\left(\begin{array}{l}0 \\ 1 \\ 1\end{array}\right)\)
  \(\overrightarrow{A C}  =\overrightarrow{O C}-\overrightarrow{O A}=\left(\begin{array}{l}3 \\ 2 \\ 4\end{array}\right)-\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 1 \\ 2\end{array}\right)\)

\(\text{Area}=\dfrac{1}{2}\ \bigg|\overrightarrow{AB} \times \overrightarrow{AC}\bigg|=\dfrac{1}{2}\ \left|\begin{array}{ccc}\underset{\sim}{i} & \underset{\sim}{j} & \underset{\sim}{k} \\ 0 & 1 & 1 \\ 2 & 1 & 2\end{array}\right|=\dfrac{1}{2}\ \bigg|\underset{\sim}{i}+2 \underset{\sim}{j}-2 \underset{\sim}{k}\bigg|=\dfrac{3}{2}\)

 
b.    \(\text {In } \triangle ABC \text {, shortest distance of } B \text { to } A C\ \text {is the} \perp \text {distance }\)

\(|\overrightarrow{A C}|= \displaystyle{\sqrt{2^2+1^2+2^2}}=3\)

\(\dfrac{3}{2}=\dfrac{1}{2}\ |\overrightarrow{A C}| \times h \ \Rightarrow \ h=1\)

\(\therefore \text { Shortest distance }=1 \text { unit }\)
 

♦♦ Mean mark (b) 35%.

c.    \(\text {Vector} \perp \text {to plane}\ \psi\  \text {is}\ \ \underset{\sim}{n}=2 \underset{\sim}{i}-2 \underset{\sim}{j}-\underset{\sim}{k}\)

\(\text{Parallel line to}\ \underset{\sim}{r}(t) \ \text{is}\ \ \underset{\sim}{m}=\underset{\sim}{i}-2 \underset{\sim}{j}+2\underset{\sim}{k}\)

\(\text{Find angle } \alpha \text{ between }\underset{\sim}{n} \text{ and } \underset{\sim}{m},\)

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

\((2 \underset{\sim}{i}-2 \underset{\sim}{j}-\underset{\sim}{k})(\underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{2 k})=3 \times 3 \times \cos \, \alpha\)

\(\Rightarrow \alpha=64^{\circ} \text { (nearest degree) }\)
 

\(\text{Find angle } \theta \text{ between } \underset{\sim}{m} \text{ and plane } \psi :\)

\(\theta=90-64=26^{\circ}\)
 

d.    \(\underset{\sim}{r}(t)=\underset{\sim}{n} \times t=2 t \underset{\sim}{i}-2 t \underset{\sim}{j}-t \underset{\sim}{k}\)

\(x=2 t, y=-2 t, z=-t \quad \text{(also accepted)}\)
 

♦ Mean mark (d) 52%.

e.    \(|\overrightarrow{OD}|=\text{shortest distance from } O \text{ to plane } \psi\).

\(\Rightarrow D \text{ is on } L \text{ and } \psi\)

\(\text{Solve for } t: \  2(2 t)-2(-2 t)-(t)=-18\)

\(\Rightarrow t=-2\)

\(|\overrightarrow{OD}|=-4 \underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{k}\)

\(\text {Shortest distance }=\displaystyle{\sqrt{(-4)^2+4^2+2^2}}=6\)
 

f.    \(\overrightarrow{OD}=-4 \underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{i}\)

\(D(-4,4,2)\)

♦ Mean mark (f) 41%.

PHYSICS, M6 2021 VCE 5

Figure 5 shows a stationary electron (e\(^{-}\)) in a uniform magnetic field between two parallel plates. The plates are separated by a distance of 6.0 × 10\(^{-3}\) m, and they are connected to a 200 V power supply and a switch. Initially, the plates are uncharged. Assume that gravitational effects on the electron are negligible.
 

  1. Explain why the magnetic field does not exert a force on the electron. Justify your answer with an appropriate formula.  (2 marks)

The switch is now closed.

  1. Determine the magnitude and the direction of any electric force now acting on the electron. Show your working.  (3 marks)

  1. Ravi and Mia discuss what they think will happen regarding the size and the direction of the magnetic force on the electron after the switch is closed.

    Ravi says that there will be a magnetic force of constant magnitude, but it will be continually changing direction.

    Mia says that there will be a constantly increasing magnetic force, but it will always be acting in the same direction.

    Evaluate these two statements, giving clear reasons for your answer.  (4 marks)

Show Answers Only

a.    → Force of a magnetic field on an electron  \(\Rightarrow F=qvB\).

→ Electron is stationary (\(v=0\))

→ The force on the electron = \(0\).

b.    \(F=qE=\dfrac{qV}{d}=\dfrac{1.602 \times 10^{-19} \times 200}{6 \times 10^{-3}}=5.34 \times 10^{-15}\ \text{N}\)

→ Since the bottom plate is positively charged, the direction of the electron will be down the page (towards the bottom plate).

c.    Both statements contain a correct assertion and incorrect assertion.

→ Ravi was incorrect to say that the magnitude of the magnetic force on the electron will be constant.

→ Mia, however, was correct in saying that the magnetic force on the electron will increase in magnitude.

→ This is due to the velocity of the electron increasing under the acceleration of the electric field. As the velocity of the electron increases, so will the magnitude of the force of the magnetic field on the electron (\(F=qvB\)). 

→ Mia was incorrect to say that the direction of the magnetic force on the electron would not change.

→ Instead, Ravi was correct in saying that the magnetic force would be continuously changing direction.

→ This is because the force of the magnetic field on the electron will act perpendicular to the velocity of the electron and will cause the electron to move in circular motion. As the force is always perpendicular to the velocity of the electron, it will continually change to act towards the centre of the electrons circular motion.

Show Worked Solution

a.    → Force of a magnetic field on an electron  \(\Rightarrow F=qvB\).

→ Electron is stationary (\(v=0\))

→ The force on the electron = \(0\).
 

♦♦ Mean mark (a) 35%. 
COMMENT: Many students confused the forces of electric (not applicable here) and magnetic fields.

b.    \(F=qE=\dfrac{qV}{d}=\dfrac{1.602 \times 10^{-19} \times 200}{6 \times 10^{-3}}=5.34 \times 10^{-15}\ \text{N}\)

→ Since the bottom plate is positively charged, the direction of the electron will be down the page (towards the bottom plate).
 

c.    Both statements contain a correct assertion and incorrect assertion.

→ Ravi was incorrect to say that the magnitude of the magnetic force on the electron will be constant.

→ Mia, however, was correct in saying that the magnetic force on the electron will increase in magnitude.

→ This is due to the velocity of the electron increasing under the acceleration of the electric field. As the velocity of the electron increases, so will the magnitude of the force of the magnetic field on the electron (\(F=qvB\)). 

→ Mia was incorrect to say that the direction of the magnetic force on the electron would not change.

→ Instead, Ravi was correct in saying that the magnetic force would be continuously changing direction.

→ This is because the force of the magnetic field on the electron will act perpendicular to the velocity of the electron and will cause the electron to move in circular motion. As the force is always perpendicular to the velocity of the electron, it will continually change to act towards the centre of the electrons circular motion.

♦♦♦ Mean mark (c) 18%.
COMMENT: Students need well planned responses to properly explain the physics principles required for 4 marks.