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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.

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 33

Explain why the boiling points of hydrogen halides \(\ce{(HF, HCl, HBr}\), and \(\ce{HI)}\) increase from \(\ce{HF}\) to \(\ce{HI}\).

Include in your answer the types of intermolecular forces involved and how molecular mass affects these boiling points.   (3 marks)

Show Answers Only

→ \(\ce{HF}\) exhibits hydrogen bonding while \(\ce{HCl, HBr}\), and \(\ce{HI}\) experience dipole-dipole interactions and dispersion forces.

→ As molecular mass increases from \(\ce{HCl}\) to \(\ce{HI}\), the number of electrons increases, enhancing the dispersion forces.

→ Despite \(\ce{HF}\) having hydrogen bonds, the significantly greater molecular mass of \(\ce{HI}\) results in stronger dispersion forces that lead to a higher boiling point.

Show Worked Solution

→ \(\ce{HF}\) exhibits hydrogen bonding while \(\ce{HCl, HBr}\), and \(\ce{HI}\) experience dipole-dipole interactions and dispersion forces.

→ As molecular mass increases from \(\ce{HCl}\) to \(\ce{HI}\), the number of electrons increases, enhancing the dispersion forces.

→ Despite \(\ce{HF}\) having hydrogen bonds, the significantly greater molecular mass of \(\ce{HI}\) results in stronger dispersion forces that lead to a higher boiling point.

Calculus, SPEC2 2022 VCAA 3

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

  1.  i. Express the differential equation in the form \(\displaystyle \int g(x)dx=\int f(t)dt\).   (1 mark)
  2. ii. Hence, show that  \(x=\log _e\left(\tan ^{-1}(2 t)+1\right)\).   (2 marks)
  3. The graph of  \(x=\log _e\left(\tan ^{-1}(2 t)+1\right)\) has a horizontal asymptote.
    1. Write down the equation of this asymptote.   (1 mark)
    2. Sketch the graph of  \(x=\log _e\left(\tan ^{-1}(2 t)+1\right)\) and the horizontal asymptote on the axes below. Using coordinates, plot and label the point where  \(t=10\), giving the value of \(x\) correct to two decimal places.   (2 marks)
       

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

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

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

  1. Verify that the particles are the same distance from \(O\) when  \(t=6\).   (1 mark)
  2. Find the ratio of the speed of the first particle to the speed of the second particle when the particles are at the same distance from \(O\). Give your answer as \(\dfrac{a}{b}\) in simplest form, where \(a\) and \(b\) are positive integers.   (2 marks)
Show Answers Only

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

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

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

b.ii. 
           

c.    \(0.02\)

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

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

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

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

Show Worked Solution

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

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

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

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

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

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

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

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

♦♦♦ Mean mark 23%.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

Functions, MET2 2022 VCAA 4

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

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

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

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

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

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

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

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

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

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

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

a.   `R` is the range.

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

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

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

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

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

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

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

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

`\frac{1}{k}(4)<1`   [Using CAS]

`:.\  k > 4`


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

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

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


♦♦♦♦ Mean mark (e.ii) 10%.

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.

Calculus, SPEC2 2022 VCAA 14 MC

A particle moving in a straight line with constant acceleration has a velocity of 7 ms\(^{-1}\) at point \(A\) and 17 ms\(^{-1}\) at point \(B\).

The velocity of the particle, in metres per second, at the midpoint of \(AB\) is

  1. \(\sqrt{119}\)
  2. \(11\)
  3. \(12\)
  4. \(13\)
  5. \(\sqrt{240}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Using}\ \ v^2=u^2+2as:\)

\(17^2\) \(=7^2+2as\)  
\(as\) \(=120\)  

 
\(\text{Displacement (halfway)}\ = \dfrac{s}{2} \)

\(v^2\) \(=7^2+2a \times \dfrac{s}{2} \)  
  \(=49+120\)  
  \(=169\)  
\(\therefore v\) \(=13\)  

 
\(\Rightarrow D\)

♦♦♦ Mean mark 28%.

PHYSICS, M6 2020 VCE 7 MC

An ideal transformer has an input DC voltage of 240 V, 2000 turns in the primary coil and 80 turns in the secondary coil.

The output voltage is closest to

  1. 0 V
  2. 9.6 V
  3. 6.0 × 10\(^{3}\) V
  4. 3.8 × 10\(^{7}\) V
Show Answers Only

\(A\)

Show Worked Solution

→ Transformers require an AC current to provide the change in flux that generates an EMF in the secondary coil from the primary coil.

→ The question refers to a DC current. There will therefore be no change in flux in the transformer and no output voltage.

\(\Rightarrow A\)

♦♦♦ Mean mark 16%.
COMMENT: Attention! DC current will not work on a transformer.

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.

Probability, MET2 2022 VCAA 20 MC

A soccer player kicks a ball with an angle of elevation of `theta` °, where `theta` is a normally distributed random variable with a mean of 42° and a standard deviation of 8°.

The horizontal distance that the ball travels before landing is given by the function `d=50 \ sin (2\theta)`.

The probability that the ball travels more than 40 m horizontally before landing is closest to

  1. 0.969
  2. 0.937
  3. 0.226
  4. 0.149
  5. 0.027
Show Answers Only

`A`

Show Worked Solution

Using CAS

Solve:  `50\ sin(2\theta)>=40`   for    `0<= theta <= 90`°

`26.565 \ldots \leq \theta \leq 63.434 \ldots`
 

`\theta \~ N\left(42°,8^2°\right)`

`\text{Pr}(26.565 \leq \theta \leq 63.435)~~0.96947`
 
`=>A`


♦♦ Mean mark 30%.
MARKER’S COMMENT: 24% incorrectly chose B and 25% incorrectly chose C.

Calculus, SPEC2 2022 VCAA 10 MC

Consider the curve given by  `5 x^2 y-3 x y+y^2=10`.

The equation of the tangent to this curve at the point `(1, m)`, where `m` is a real constant, will have a negative gradient when

  1. `m \in R \backslash[-1,0]`
  2. `m=-\sqrt{11}-1 \ text {only}`
  3. `m \in R \backslash(-1,0]`
  4. `m=\sqrt{11}-1 \ text[only]`
  5. `m=-\sqrt{11}-1 or m=\sqrt{11}-1`
Show Answers Only

`E`

Show Worked Solution

`5 x^2 y-3 x y+y^2=10`

`text{At}\ (1,m):\ \ 5m-3m+m^2=10`

`text{Solve (by CAS):}\ \ m=-1+-sqrt{11}`

`text{Check the value of the derivative function at both values of}\ m\ (x=1)`

`text{By CAS, both values (gradients) are negative.}`

`=>E`

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.

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.

PHYSICS, M6 2021 VCE 2

A schematic side view of one design of an audio loudspeaker is shown in Figure 2. It uses a current carrying coil that interacts with permanent magnets to create sound by moving a cone in and out.
 

Figure 3 shows a schematic view of the loudspeaker from the position of the eye shown in Figure 2. The direction of the current is clockwise, as shown.
 

   

  1. Draw four magnetic field lines on Figure 3, showing the direction of each field line using an arrow.  (1 mark)

  2. Which one of the following gives the direction of the force acting on the current carrying coil shown in Figure 3?  (1 mark)
A. left B. right
C. up the page D. down the page
E. into the page F. out of the page
  1. The current carrying coil has a radius of 5.0 cm and 20 turns of wire, and it carries a clockwise current \((I)\) of 2.0 A. Its magnetic field strength \((B)\) is 200 mT.
  2. Calculate the magnitude of the force, \(F\), acting on the current carrying coil. Show your working.  (2 marks)

Show Answers Only

a.   
         

b.    \(E\)

c.    \(2.51\ \text{N}\)

Show Worked Solution

a.    Magnetic fields run from the north pole to the south pole.
 

   

 

b.    Apply the right-hand rule:

→ Thumb to the right and fingers down, the force on the current carrying coil must be into the page.

\(\Rightarrow E\)
 

c.    \(l= 2\pi \times 5 \times 10^{-2} = 0.1\pi\)

\(\therefore F=nlIB=20 \times 0.1\pi \times 2 \times 0.2=2.51\ \text{N}\)

♦♦♦ Mean mark 24%.

Functions, MET2 2023 VCAA 2

The following diagram represents an observation wheel, with its centre at point \(P\). Passengers are seated in pods, which are carried around as the wheel turns. The wheel moves anticlockwise with constant speed and completes one full rotation every 30 minutes.When a pod is at the lowest point of the wheel (point \(A\)), it is 15 metres above the ground. The wheel has a radius of 60 metres.
 

Consider the function \(h(t)=-60\ \cos(bt)+c\) for some \(b, c \in R\), which models the height above the ground of a pod originally situated at point \(A\), after time \(t\) minutes.

  1. Show that \(b=\dfrac{\pi}{15}\) and \(c=75\).   (2 marks)
  2. Find the average height of a pod on the wheel as it travels from point \(A\) to point \(B\).
  3. Give your answer in metres, correct to two decimal places.   (2 marks)
  4. Find the average rate of change, in metres per minute, of the height of a pod on the wheel as it travels from point \(A\) to point \(B\).   (1 mark)

After 15 minutes, the wheel stops moving and remains stationary for 5 minutes. After this, it continues moving at double its previous speed for another 7.5 minutes.

The height above the ground of a pod that was initially at point \(A\), after \(t\) minutes, can be modelled by the piecewise function \(w\):
 

\(w(t) = \begin {cases}
h(t)         &\ \ 0 \leq t < 15 \\
k         &\ \ 15 \leq t < 20 \\
h(mt+n) &\ \ 20\leq t\leq 27.5
\end{cases}\)

 
where \(k\geq 0, m\geq 0\) and \(n \in R\).

    1. State the values of \(k\) and \(m\).   (1 mark)
    1. Find all possible values of \(n\).  (2 marks)
    2. Sketch the graph of the piecewise function \(w\) on the axes below, showing the coordinates of the endpoints.   (3 marks)

       
Show Answers Only

a.    \(\text{See worked solution}\)

b.    \(\approx 36.80\ \text{m}\)

c.    \(8\)

d.i.  \(k=135, m=2\)

d.ii. \(n=30p+5,\ p\in Z\)

d.iii. 

Show Worked Solution
a.    \(\text{Period:}\) \(\dfrac{2\pi}{b}\) \(=30\)
    \(\therefore\ b\) \(=\dfrac{\pi}{15}\)

  
\(\text{Given }h(t)=-60\cos(bt)+c,\ \text{evaluate when }t=0, h=15\ \text{to find }c.\)

\(-60\ \cos(0)+c\)  \(=15\)
\(c\) \(=75\)

  \(\therefore\ h(t)=-60\cos(\dfrac{\pi t}{15})+75\)

b.     \(\text{Average height}\) \(=\dfrac{1}{\frac{30}{4}-0}\displaystyle\int_0^{\frac{30}{4}}\Bigg(-60\cos\Bigg(\dfrac{\pi}{15}t\Bigg)+75\Bigg)\,dt\)
    \(=\dfrac{2}{15}\left[75t-\dfrac{900\ \sin(\frac{\pi}{15}t)}{\pi}\right]_{0}^{7.5}\)
    \(=\dfrac{2}{15}\Bigg[75\times 7.5-\dfrac{900\ \sin(\frac{\pi}{15}\times 7.5)}{\pi}\Bigg]-\Bigg[0\Bigg]\)
    \(=\dfrac{75\pi-120}{\pi}=\dfrac{15(5\pi-8)}{\pi}\)
    \(=36.802\dots\approx 36.80\ \text{m}\)
 
♦♦ Mean mark (b) 45%.
MARKER’S COMMENT: \(\frac{1}{60}\int_0^{60} h(t)dt\) was a common error.
Others incorrectly found average rate of change instead of average value.
c.     \(\text{Av rate of change of height}\) \(=\dfrac{h(7.5)-h(0)}{7.5}\)
    \(=\dfrac{\Bigg(75-60\cos(\dfrac{\pi \times 7.5}{15})\Bigg)-\Bigg(75-60\cos(\dfrac{\pi\times 0}{15})\Bigg)}{7.5}\)
    \(=\dfrac{75-15}{7.5}=8\)
 
♦ Mean mark (c) 50%.
MARKER’S COMMENT: Common incorrect answer was – 8.
di.    \(\text{Period is 30 minutes, so after 15 minutes pod}\)
\(\text{is at the top of the wheel.}\)
  \(\therefore\ k=75-60\ \cos(\frac{\pi}{15}\times 15)=135\)

  
\(\text{Pod is travelling at twice its previous speed}\)
\(\text{so one revolution takes 15 minutes}\)

\(\therefore\ \text{Period}\rightarrow\) \(\dfrac{2\pi}{bm}\) \(=15\)
  \(bm\) \(=\dfrac{2\pi}{15}\)
  \(\dfrac{\pi}{15}\cdot m\) \(=\dfrac{2\pi}{15}\)
  \(m\) \(=\dfrac{2\pi}{15}\times \dfrac{15}{\pi}\)
    \(=2\)
 
♦♦ Mean mark (d)(i) 40%.
MARKER’S COMMENT: Many students could find \(k\) but not \(m\) with \(m=\frac{1}{2}\) a common error.
dii.   \(\text{When }t=20\ \text{the pod is at the top of the wheel and height is 135.}\)
  
\(\text{When }t=27.5\ \text{the pod is back at the start and height is 15.}\)

   
\(\text{Using CAS solve for }t=20:\rightarrow\ h(2(20)+n)=135\ \rightarrow\ n=30p+5\)

\(\text{Using CAS solve for }t=27.5:\rightarrow\ h(2(27.5)+n)=15\ \rightarrow\ n=30p+5\)

\(\therefore\ n=30p+5,\ p\in Z\)

 
♦♦♦ Mean mark (d)(ii) 20%.
MARKER’S COMMENT: Many students set up the correct equations to solve but did not provide the general solution. Some incorrectly stated the variable as an element of R.

diii.   

 
♦♦♦ Mean mark (d)(iii) 40%.
MARKER’S COMMENT: Students are reminded to include all endpoints and be particular about the curvature of graph.

Calculus, MET2 2023 VCAA 1

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

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

a.    \((-1, 0), (0, 0), (2, 0)\)

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

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

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

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

c.iii. \(5.95\)

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

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

Show Worked Solution

a.    \((-1, 0), (0, 0), (2, 0)\)
  

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

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

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

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

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

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

ci    \(\text{Given }f(x)=g(x)\)

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

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

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

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

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

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

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

  

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

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

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

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

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

\(2ab+b^2=-2\ \ …(2)\)

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

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

  

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

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

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

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

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

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

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

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

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

 

PHYSICS, M5 2022 VCE 2

There are over 400 geostationary satellites above Earth in circular orbits. The period of orbit is one day (86 400 seconds). Each geostationary satellite remains stationary in relation to a fixed point on the equator. Figure 2 shows an example of a geostationary satellite that is in orbit relative to a fixed point, \(\text{X}\), on the equator.
 

  1. Explain why geostationary satellites must be vertically above the equator to remain stationary relative to Earth's surface.  (2 marks)
  1.  Using  \(G=6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}, M_{\text{E}} = 5.98 \times 10^{24}\ \text{kg}\)  and  \(R_{\text{E}} = 6.37 \times 10^{6}\ \text{m}\), show that the altitude of a geostationary satellite must be equal to \(3.59 \times 10^{7}\ \text{m}\).  (4 marks)
  1. Calculate the speed of an orbiting geostationary satellite.  (3 marks)

Show Answers Only

a.   Reasons the satellite must orbit relative to a fixed point on the equator:

→ So that it is in the same rotational plane/axis of the Earth. 

→ The gravitational force on the satellite must be directly towards the centre of the Earth and it must orbit at a set altitude to the Earth where the speed of the satellite matches the rotational speed of the Earth.

→ As the speed and direction of the satellite matches that of the Earth, it will remain stationary relative to the motion of the Earth.

b.    See worked solutions

c.    3076 ms\(^{-1}\)

Show Worked Solution

a.   Reasons the satellite must orbit relative to a fixed point on the equator:

→ So that it is in the same rotational plane/axis of the Earth. 

→ The gravitational force on the satellite must be directly towards the centre of the Earth and it must orbit at a set altitude to the Earth where the speed of the satellite matches the rotational speed of the Earth.

→ As the speed and direction of the satellite matches that of the Earth, it will remain stationary relative to the motion of the Earth.
 

♦♦♦ Mean mark (a) 19%.
b.     \(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)
  \(r\) \(=\sqrt[3]{\dfrac{GMT^2}{4\pi^2}}\)
    \(=\sqrt[3]{\dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times (86\ 400)^2}{4\pi^2}}\)
    \(=42.250 \times 10^6\ \text{m}\)

 
\(\therefore \ \text{Altitude}\ =42.250 \times 10^6 -6.37 \times 10^6 =3.59 \times 10^7\ \text{m}\)
 

c.     \(v\) \(=\sqrt{\dfrac{GM}{r}}\)
    \(=\sqrt{\dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{42.25 \times 10^6}}\)
    \(=3073\ \text{ms}^{-1}\)

Complex Numbers, SPEC2 2023 VCAA 2

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

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

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

    1. Find the equation of this ray in the form \(\text{Arg}\left(z-z_0\right)=\theta\), where \(z_0 \in C\), and \(\theta\) is measured in radians in terms of \(\pi\).  (1 mark)
  1. Verify that the equation  \(z^7-1=0\)  can be expressed in the form  \((z-1)\left(z^6+z^5+z^4+z^3+z^2+z+1\right)=0\).
  2.  
    1. Express  \(\text{cis}\left(\dfrac{2 \pi}{7}\right)+\operatorname{cis}\left(\dfrac{12 \pi}{7}\right)\) in the form \(A \cos (B \pi)\), where \(A, B \in R^{+}\).
    2. Given that  \(w=\operatorname{cis}\left(\dfrac{2 \pi}{7}\right)\)  satisfies  \((z-1)\left(z^6+z^5+z^4+z^3+z^2+z+1\right)=0\), use De Moivre's theorem to show that \(\cos \left(\dfrac{2 \pi}{7}\right)+\cos \left(\dfrac{4 \pi}{7}\right)+\cos \left(\dfrac{6 \pi}{7}\right)=-\dfrac{1}{2}\).

Show Answers Only

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

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

b.  \(\text{Roots of}\ \ z^7-1=0:\) 

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

c.   
          

d.i.   
          

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

\(\arg(z-1)=\dfrac{9\pi}{14} \)
 

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

 

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

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

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

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

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

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

Show Worked Solution

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

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

b.  \(\text{Roots of}\ \ z^7-1=0:\) 

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

c.   
          

d.i.   
          

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

\(\arg(z-1)=\dfrac{9\pi}{14} \)
 

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

 

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

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

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

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

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

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

Probability, MET2 2023 VCAA 4

A manufacturer produces tennis balls.

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

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

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

  1. Find the probability that a randomly selected tennis ball can fit through the opening at the top of the container.
    Give your answer correct to four decimal places.   (1 mark)

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

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

  1. Given that a tennis ball can fit through the opening at the top of the container, find the probability that it is classed as grade A.
    Give your answer correct to four decimal places.   (2 marks)

  2. The manufacturer would like to improve processes to ensure that more than 99% of all tennis balls produced are classed as grade A.
     
    Assuming that the mean diameter of the tennis balls remains the same, find the required standard deviation of the diameter, in centimetres, correct to two decimal places.   (2 marks)

  3. An inspector takes a random sample of 32 tennis balls from the manufacturer and determines a confidence interval for the population proportion of grade A balls produced.
    The confidence interval is (0.7382, 0.9493), correct to four decimal places.
     
    Find the level of confidence that the population proportion of grade A balls is within the interval, as a percentage correct to the nearest integer.   (2 marks)

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

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

  1. Find the probability that the serving speed of a grade A ball exceeds 50 metres per second.
    Give your answer correct to four decimal places.   (1 mark)

  2. Find the exact mean serving speed for grade A balls, in metres per second.   (1 mark)

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

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

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

a.    \(0.1587\)

b.    \(d\approx6.83\)

c.    \(0.9938\)

d.    \(0.9998\)

e.    \(0,8960\)

f.    \(0.06\)

g.    \(90\%\)

h.    \(0.1345\)

i.    \(3(\pi^2+4)\)

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

Show Worked Solution

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

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

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

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

b.    \(\Pr(D<d)=0.90\)  

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

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

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

 

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

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

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

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

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

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

\(X\sim \text{Bi}(4, 0.999379 …)\)

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

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

 

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

  

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

 

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

♦♦♦ Mean mark (j) 10%.

Calculus, MET2 2023 VCAA 5

Let \(f:R \to R, f(x)=e^x+e^{-x}\) and \(g:R \to R, g(x)=\dfrac{1}{2}f(2-x)\).

  1. Complete a possible sequence of transformations to map \(f\) to \(g\).   (2 marks)
    •    Dilation of factor \(\dfrac{1}{2}\) from the \(x\) axis.

Two functions \(g_1\) and \(g_2\) are created, both with the same rule as \(g\) but with distinct domains, such that \(g_1\) is strictly increasing and \(g_2\) is strictly decreasing.

  1. Give the domain and range for the inverse of \(g_1\).   (2 marks)

Shown below is the graph of \(g\), the inverse of \(g_1\) and \(g_2\), and the line \(y=x\).
 

The intersection points between the graphs of \(y=x, y=g(x)\) and the inverses of \(g_1\) and \(g_2\), are labelled \(P\) and \(Q\).

    1. Find the coordinates of \(P\) and \(Q\), correct to two decimal places.   (1 mark)
    1. Find the area of the region bound by the graphs of \(g\), the inverse of \(g_1\) and the inverse of \(g_2\).
      Give your answer       correct to two decimal places.   (2 marks)

Let \(h:R\to R, h(x)=\dfrac{1}{k}f(k-x)\), where \(k\in (o, \infty)\).

  1. The turning point of \(h\) always lies on the graph of the function \(y=2x^n\), where \(n\) is an integer.
    Find the value of \(n\).  (1 mark)

Let \(h_1:[k, \infty)\to R, h_1(x)=h(x)\).

The rule for the inverse of \(h_1\) is \(y=\log_{e}\Bigg(\dfrac{1}{k}x+\dfrac{1}{2}\sqrt{k^2x^2-4}\Bigg)+k\)

  1. What is the smallest value of \(k\) such that \(h\) will intersect with the inverse of \(h_1\)?
    Give your answer correct to two decimal places.   (1 mark)

It is possible for the graphs of \(h\) and the inverse of \(h_1\) to intersect twice. This occurs when \(k=5\).

  1. Find the area of the region bound by the graphs of \(h\) and the inverse of \(h_1\), where \(k=5\).
    Give your answer correct to two decimal places.   (2 marks)

Show Answers Only

a.    \(\text{Reflect in the }y\text{-axis}\)

\(\text{Then translate 2 units to the right}\)

\(\text{OR}\)

\(\text{Translate 2 units to the left}\)

\(\text{Then reflect in the }y\text{-axis}\)

\(\text{OR}\)

\(\text{Translate 2 units to the right only }(f \ \text{even function})\)

b.    \(\text{See worked solution}\)

c.i.  \(P(1.27, 1.27\), Q(4.09, 4.09)\)

c.ii. \(2\displaystyle \int_{1.27…}^{4.09…} (x-g(x))\,dx=5.56\)

d.    \(n=-1\)

e. \(k\approx 1.27\)

f. \(A=43.91\)

Show Worked Solution

a.    \(\text{Reflect in the }y\text{-axis}\)

\(\text{Then translate 2 units to the right}\)

\(\text{OR}\)

\(\text{Translate 2 units to the left}\)

\(\text{Then reflect in the }y\text{-axis}\)

\(\text{OR}\)

\(\text{Translate 2 units to the right only }(f \ \text{even function})\)

b.    \(\text{Some options include:}\)

•  \(\text{Domain}\to [1, \infty)\ \ \ \text{Range}\to [2, \infty)\)
\(\text{or}\)
•  \(\text{Domain}\to (1, \infty)\ \ \ \text{Range}\to (2, \infty)\)
 

\(\text{If functions split at turning point } (2, 1)\ \text{then possible options:}\)

• \(\text{Domain}\ g_1\to [2, \infty)\ \ \ \text{Range}\to [1, \infty)\)
   \(\text{and }\)
  \(\text{Domain}\ g_1^{-1}\to [1, \infty)\ \ \ \text{Range}\to [2, \infty)\)

\(\text{or}\)

•  \(\text{Domain}\ g_1\to (2, \infty)\ \ \ \text{Range}\to (1, \infty)\)
   \(\text{and }\)
   \(\text{Domain}\ g_1^{-1}\to (1, \infty)\ \ \ \text{Range}\to (2, \infty)\)


♦ Mean mark 50%.
MARKER’S COMMENT: Note: maximal domain not requested so there were many correct answers. Many students seemed to be confused by notation. Often domain and range reversed.

c.i.  \(\text{By CAS:}\  P(1.27, 1.27), Q(4.09, 4.09)\)

c.ii.   \(\text{Area}\) \(=2\displaystyle \int_{1.27…}^{4.09…} (x-g(x))\,dx\)
    \(=2\displaystyle \int_{1.27…}^{4.09…} \Bigg(x-\dfrac{1}{2}\Big(e^{2-x}+e^{x-2}\Big)\Bigg)\,dx\)
    \(\approx 5.56\)

♦♦ Mean mark (c)(ii) 30%.
MARKER’S COMMENT: Some students set up the integral but did not provide an answer.
Others did not double the integral. Some incorrectly used \(\displaystyle \int_{1.27…}^{4.09…}(g_1^-1-g(x))\,dx\).
Others had correct answer but incorrect integral.

d.    \((0, 2)\ \text{turning pt of }f(x)\)

\(\text{and }h(x) \text{is the transformation from }f(x)\)

\(\therefore \Bigg(k, \dfrac{2}{k}\Bigg)\ \text{turning tp of }h(x)\)

\(\text{so the coordinates have the relationship}\)

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

\(\therefore\ n=-1\)


♦♦♦ Mean mark (d) 10%.
MARKER’S COMMENT: Question not well done. \(n=1\) a common error.

e.    \(\text{Smallest }k\ \text{is when inverse of the curves }h_1\ \text{and}\ h\)

\(\text{touch with the line }y=x.\)

\(\therefore\ h(x)\) \(=\dfrac{1}{k}\Big(e^{k-x}+e^{x-k}\Big)\)
  \(=x\)

 

\(\text{and}\ h'(x)\) \(=\dfrac{1}{k}\Big(-e^{k-x}+e^{x-k}\Big)\)
  \(=1\)

\(\text{at point of intersection, where }k>0.\)

\(\text{Using CAS graph both functions to solve or solve as simultaneous equations.}\)

\(\rightarrow\ k\approx 1.2687\approx 1.27\)


♦♦♦♦ Mean mark (e) 0%.
MARKER’S COMMENT: Question poorly done with many students not attempting.

f.    \(\text{When}\ k=5\)

\(h_1^{-1}(x)=\log_e\Bigg(\dfrac{5}{2}x+\dfrac{1}{2}\sqrt{25x^2-4}\Bigg)+5\)

\(\text{and }h(x)=\dfrac{1}{5}\Big(e^{5-x}+e^{x-5}\Big)\)

\(\text{Using CAS: values of }x\ \text{at of intersection of fns are}\)

\(x\approx 1.45091…\ \text{and}\ 8.78157…\)

\(\text{Area}\) \(=\displaystyle \int_{1.45091…}^{8.78157…}h_1^{-1}(x)- h(x)\,dx\)
  \(=\displaystyle \int_{1.45091…}^{8.78157…} \log_e\Bigg(\dfrac{5}{2}x+\dfrac{1}{2}\sqrt{25x^2-4}\Bigg)+5-\dfrac{1}{5}\Big(e^{5-x}+e^{x-5}\Big)\,dx\)
  \(=43.91\) 

♦♦♦ Mean mark (f) 15%.
MARKER’S COMMENT: Errors were made with terminals. Common error using \(x=2.468\), which was the \(x\) value of the pt of intersection of \(h\) and \(y=x\).

Calculus, MET2 2023 VCAA 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. \(-10\)
  2. \(-6\)
  3. \(0\)
  4. \(6\)
  5. \(10\)
Show Answers Only

\(E\)

Show Worked Solution

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

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

 
\(\Rightarrow E\)


♦♦♦ Mean mark 22%.
MARKER’S COMMENT: 51% of students incorrectly chose D (6).

Calculus, MET1 2022 VCAA 8

Part of the graph of `y=f(x)` is shown below. The rule `A(k)=k \ sin(k)` gives the area bounded by the graph of `f`, the horizontal axis and the line `x=k`.
 

  1. State the value of `A\left(\frac{\pi}{3}\right)`.   (1 mark)

  2. Evaluate `f\left(\frac{\pi}{3}\right)`.   (2 marks)

  3. Consider the average value of the function `f` over the interval `x \in[0, k]`, where `k \in[0,2]`.
  4. Find the value of `k` that results in the maximum average value.   (2 marks)

Show Answers Only

a.    ` (sqrt(3)pi)/6`

b.    `(3sqrt(3) +pi)/6`

c.    `k = pi/2`

Show Worked Solution
a.  `A(pi/3)` `= pi/3sin(pi/3)`  
  `= pi/3 xx sqrt(3)/2`  
  `= (sqrt(3)pi)/6`  
b.   `f(k)` `= A^{\prime}(k)`  
`f(k)` `=d/dx(k\ sin\ k)`  
  `= sin\ k + k\ cos\ k`  
`f(pi/3)` `= sin\ pi/3 + pi/3\ cos\ pi/3`  
  `= sqrt(3)/2 + pi/3 xx 1/2`  
  `= sqrt(3)/2 + pi/6`  
  `=(3sqrt(3) +pi)/6`  

♦♦♦ Mean mark (b) 25%.
MARKER’S COMMENT: Common error was giving derivative of `A(k)` as `k\ cos (k)`.
c.   Average value `= (A(k))/k`  
  `= 1/k(int_0^k f(x) d x)`  
  `= 1/k[x\ sin (x)]_0^k`  
  `= 1/k[k\ sin (k)] = sin\ k`  

 
`:.` Average value has a maximum value of 1 when `k = pi/2`


♦♦♦ Mean mark (c) 25%.
MARKER’S COMMENT: Students often chose unnecessarily complicated methods or inconsistently applied nomenclature when attempting this question.

Calculus, MET1 2022 VCAA 7

A tilemaker wants to make square tiles of size 20 cm × 20 cm.

The front surface of the tiles is to be painted with two different colours that meet the following conditions:

  • Condition 1 - Each colour covers half the front surface of a tile.
  • Condition 2 - The tiles can be lined up in a single horizontal row so that the colours form a continuous pattern.

An example is shown below.
 

There are two types of tiles: Type A and Type B.

For Type A, the colours on the tiles are divided using the rule `f(x)=4 \sin \left(\frac{\pi x}{10}\right)+a`, where `a \in R`.

The corners of each tile have the coordinates (0,0), (20,0), (20,20) and (0,20), as shown below.
 

  1.  i. Find the area of the front surface of each tile.   (1 mark)

    ii. Find the value of `a` so that a Type A tile meets Condition 1.   (1 mark)


Type B tiles, an example of which is shown below, are divided using the rule `g(x)=-\frac{1}{100} x^3+\frac{3}{10} x^2-2 x+10`.
 

  1. Show that a Type B tile meets Condition 1.   (3 marks)

  2. Determine the endpoints of `f(x)` and `g(x)` on each tile. Hence, use these values to confirm that Type A and Type B tiles can be placed in any order to produce a continuous pattern in order to meet Condition 2.   (2 marks)

Show Answers Only

a.i.    `400` cm²

a.ii.   `a = 10`

b.     `200` cm²

c.     See worked solution

Show Worked Solution
a.i   Area `= 20 xx 20`  
  `= 400` cm²  

  
a.ii  `a = 10`
 

b.   Area `=\int_0^{20} \frac{-x^3}{100}+\frac{3 x^2}{10}-2 x+10\ d x`  
  `=\left[\frac{-x^4}{400}+\frac{x^3}{10}-x^2+10 x\right]_0^{20}`  
  `=\left[-\frac{20^4}{400}+\frac{20^3}{10}-20^2+10 xx 20\right]-\left[0\right]`  
  `= – 400 +800 -400 +200`  
  `= 200` cm²  

 
`:.` The area of the coloured section of the Type B tile is 200 cm² which is half the 400 cm² area of the tile.
 

c.   
`f(0)` `=4 \sin \left(\frac{\pi(0)}{10}\right)+10 = 10`  
  `f(20)` `=4 \sin (2 \pi)+10 = 10`  
  `g(0)` `=\frac{-0}{100}+\frac{3(0)}{10}-2(0)+10=10`  
  `g(20)` `=-\frac{8000}{100}+\frac{200}{10}-2(20)+10 = 10`  

 
→`\ f(0) = f(20) = g(0) = g(20) =10`
 

`:.`    The endpoints for `f(x)` are `(0,10)` and `(20,10)` and for `g(x)` are also `(0,10)` and `(20,10)`.

So the tiles can be placed in any order to make the continuous pattern.


♦♦ Mean mark (c) 30%.
MARKER’S COMMENT: Students often only found `f(20)` and `g(20)`, however, `f(0)` and `g(0)` also needed to be found to verify the pattern match.

Graphs, MET1 2022 VCAA 6

The graph of `y=f(x)`, where `f:[0,2 \pi] \rightarrow R, f(x)=2 \sin(2x)-1`, is shown below.
 

  1. On the axes above, draw the graph of `y=g(x)`, where `g(x)` is the reflection of `f(x)` in the horizontal axis.   (2 marks)
  2. Find all values of `k` such that `f(k)=0` and `k \in[0,2 \pi]`.   (3 marks)
  3. Let `h: D \rightarrow R, h(x)=2 \sin(2x)-1`, where `h(x)` has the same rule as `f(x)` with a different domain.
  4. The graph of `y=h(x)` is translated `a` units in the positive horizontal direction and `b` units in the positive vertical direction so that it is mapped onto the graph of `y=g(x)`, where `a, b \in(0, \infty)`.
    1. Find the value for `b`.   (1 mark)

    2. Find the smallest positive value for `a`.   (1 mark)

    3. Hence, or otherwise, state the domain, `D`, of `h(x)`.   (1 mark)

Show Answers Only

a.    Graph `y=g(x)`

b.    `\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}`

c.i.    `b=2`

cii.    `=\frac{\pi}{2}`

ciii.   `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`

Show Worked Solution

a.

b.  `2 \sin (2 k)-1` `=0`       `0<=k<=2\pi`  
`sin (2 k)` `=1/2`  
`2k` `=\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{13 \pi}{6}, \frac{17 \pi}{6}`  
`k` `=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{13 \pi}{12}, \frac{17 \pi}{12}`  

 

c.i   ` 2 \sin 2(x-a)-1+b` `=-(2 \sin 2 x-1)`  
`:.\ -1+b` `=1`  
`b` `=2`  

 

c.ii  `2 \sin 2(x-a)` `=-(2 \sin 2 x-1)`  
`\sin (2 x-2 a)` `=-\sin 2 x`  
`\therefore 2 a` `=\pi`  
`a` `=\frac{\pi}{2}`  

♦ Mean mark (c.ii) 50%.
MARKER’S COMMENT: Students confused vertical and horizontal translations. Common error `a=\frac{\pi}{4}.`

c.iii The domain for `f(x)` is `[0,2pi]`

`:. \ D` is `\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]`

 
♦♦♦ Mean mark (c.iii) 10%.
MARKER’S COMMENT: Common error was translating in the wrong direction. A common incorrect answer was `\left[-\frac{\pi}{2}, \frac{5 \pi}{2}\right].`

Calculus, MET1 2023 VCAA 9

The shapes of two walking tracks are shown below.
 

 

Track 1 is described by the function  \(f(x)=a-x(x-2)^2\).

Track 2 is defined by the function  \(g(x)=12x-bx^2\).

The unit of length is kilometres.

  1. Given that \(f(0)=12\)  and  \(g(1)=9\), verify that \(a=12\)  and  \(b=-3\).   (1 mark)
  2. Verify that \(f(x)\) and \(g(x)\) both have a turning point at \(P\).
  3. Give the co-ordinates of \(P\).  (2 marks)
  4. A theme park is planned whose boundaries will form the triangle \(\Delta OAB\) where \(O\) is the origin, \(A\) is at \((k, 0)\) and \(B\) is at \((k, g(k))\), as shown below, where \(k \in (0, 4)\).
  5. Find the maximum possible area of the theme park, in km².   (3 marks)
     

Show Answers Only

a.    \(\text{See worked solution}\)

b.    \(P(2, 12)\ \text{Also see worked solution}\)

c.    \(\dfrac{128}{9}\)

Show Worked Solution
a.    \(f(0)\) \(=a-0(0-2)^2=12\)
  \(\therefore\ a\) \(=12\)

 

\(g(1)\) \(=12\times 1+b\times 1^2\)
\(12+b\) \(=9\)
\(\therefore b\) \(=-3\)

 

b.    \(f(x)\) \(=12-x(x-2)^2\)
    \(=-x^3+4x^2-4x+12\)
  \(f^{′}(x)\) \(=-(3x^2-8x+4)\)

 

\(\text{For turning point }f^{′}(x)=0:\)

\( -(3x^2-8x+4)\) \(=0\)
\(-(3x-2)(x-2)\) \(=0\)

 
\(\therefore\ x=\dfrac{2}{3}\ \text{or}\ 2\)
 

\(f(2)\) \(=12-2(2-2)^2=12\)
\(f\Big{(}\dfrac{2}{3}\Big{)}\) \(=12-\dfrac{2}{3}\Big(\dfrac{2}{3}-2\Big)^2<12\)

 
\(\therefore\ \text{Given graph of }f(x)\ \text{relevant turning point is }(2, 12).\)
 

\(g(x)=12x-3x^2\ \Rightarrow\ g^{′}(x)=12-6x\)

\(\text{For turning point}\ \ g^{′}(x)=0:\)

\(12-6x=0\ \ \Rightarrow \ x=2\)

\(g(x)\) \(=12x-3x^2\)
\(g(2)\) \(=12\times 2-3\times 2^2=12\)

 
\(\therefore\ g(x)\ \text{also has a turning point at }(2, 12).\)
 


♦ Mean mark (b) 48%.

c.   \(\text{Area of triangle }\Delta OAB\)

\(A(k)\) \(=\dfrac{1}{2}\times k\times g(k)\)
  \(=\dfrac{k}{2}\times (12k-3k^2)\)
  \(=6k^2-\dfrac{3k^3}{2}\)

 
\(\text{Max area when }A^{′}(k)=0:\)

\(A^{′}(k)=12k-\dfrac{9k^2}{2}=\dfrac{1}{2}(24k-9k^2)\)

\(\dfrac{1}{2}k(24-9k)=0\)

\(\therefore\ k=\dfrac{24}{9}=\dfrac{8}{3}\)

 
\(\text{Maximum area occurs when}\ k=\dfrac{8}{3}:\)

\(A(k)\) \(=\dfrac{1}{2}k\times (12k-3k^2)\)
\(A\bigg(\dfrac{8}{3}\bigg)\) \(=\dfrac{1}{2}\times\dfrac{8}{3}\times\bigg (12\bigg(\dfrac{8}{3}\bigg)-3\bigg(\dfrac{8}{3}\bigg)^2\bigg)\)
  \(=\dfrac{4}{3}\bigg(32-\dfrac{64}{3}\bigg)\)
  \(=\dfrac{4}{3}\bigg(\dfrac{96}{3}-\dfrac{64}{3}\bigg)\)
  \(=\dfrac{4}{3}\times\dfrac{32}{3}\)
  \(=\dfrac{128}{9}\)

♦♦♦ Mean mark (c) 27%.
MARKER’S COMMENT: Many students made arithmetic errors substituting the fractional values of \(k\) into \(A(k)\) to find max area.

Statistics, MET1 2023 VCAA 6

Let \(\hat{P}\) be the random variable that represents the sample proportion of households in a given suburb that have solar panels installed.

From a sample of randomly selected households in a given suburb, an approximate 95% confidence interval for the proportion \(p\) of households having solar panels installed was determined to be (0.04, 0.16).

  1. Find the value of \(\hat{p}\) that was used to obtain this approximate 95% confidence interval.   (1 mark)

Use \(z=2\) to approximate the 95% confidence interval.

  1. Find the size of the sample from which this 95% confidence interval was obtained.   (2 marks)
  2. A larger sample of households is selected, with a sample size four times the original sample.
    The sample proportion of households having solar panels installed is found to be the same.
  3. By what factor will the increased sample size affect the width of the confidence interval?   (1 mark)
Show Answers Only

a.    \(\hat{p}=0.1\)

b.    \(n=100\)

c.    \(\text{The confidence interval width is halved.}\)

Show Worked Solution

a.    \(\text{The span of a confidence interval is symmetrical about}\ \hat{p}.\)

\(\hat{p}=\dfrac{0.04+0.16}{2}=0.1\)


♦ Mean mark (a) 52%.
b.    \(0.06\) \(=2\times\sqrt{\dfrac{0.1\times0.9}{n}}\)
  \(0.03\) \(=\sqrt{\dfrac{0.1\times0.9}{n}}\)
  \((0.03)^2\) \(=\dfrac{0.1\times0.9}{n}\)
  \(0.0009\) \(=\dfrac{9}{100n}\)
  \(n\) \(=\dfrac{9}{100 \times 0.0009}\)
  \(n\) \(=100\)

♦♦ Mean mark (b) 38%.

c.    \(\text{C.I. width} \propto \dfrac{1}{\sqrt{n}}\)

→ \(\text{If}\ \ n\ \Rightarrow \ 4n,\ \ \text{C.I. width}\ \propto \dfrac{1}{\sqrt{4n}} = \dfrac{1}{2\sqrt{n}}\)

→ \(\text{The confidence interval is halved.}\)


♦♦♦ Mean mark (c) 23%.

Probability, MET1 2023 VCAA 8

Suppose that the queuing time, \(T\) (in minutes), at a customer service desk has a probability density function given by
 

\(f(t) = \begin {cases}
kt(16-t^2)         &\ \ 0 \leq t \leq 4 \\
\\
0 &\ \ \text{elsewhere}
\end{cases}\)

 
for some  \(K \in R\).

  1. Show that  \(k=\dfrac{1}{64}\).   (1 mark)
  2. Find  \(\text{E}(T)\).   (2 marks)
  3. What is the probability that a person has to queue for more than two minutes, given that they have already queued for one minute?   (3 marks)
Show Answers Only
a.    \(\displaystyle \int_{0}^{4} (16-t^2)\,dt\) \(=1\)
  \(k\left[8t^2-\dfrac{t^4}{4}\right]_0^4\) \(=1\)
  \(k\Bigg(8\times16-\dfrac{16\times16}{4}\Bigg)\) \(=1\)
  \(64k\) \(=1\)
  \(\therefore\ k\) \(=\dfrac{1}{64}\)

b.    \(E(T)=\dfrac{32}{15}\)

c.    \(\dfrac{16}{25}=0.64\)

Show Worked Solution
a.    \(\displaystyle \int_{0}^{4} kt(16-t^2)\,dt\) \(=1\)
  \(k\left[8t^2-\dfrac{t^4}{4}\right]_0^4\) \(=1\)
  \(k\Bigg(8\times16-\dfrac{16\times16}{4}\Bigg)\) \(=1\)
  \(64k\) \(=1\)
  \(\therefore\ k\) \(=\dfrac{1}{64}\)

♦ Mean mark (a) 43%.
b.    \(E(T)\) \(=\dfrac{1}{64}\displaystyle \int_{0}^{4} (16t^2-t^4)\,dt\)
    \(=\dfrac{1}{64}\left[\dfrac{16t^3}{3}-\dfrac{t^5}{5}\right]_0^4\)
    \(=\dfrac{1}{64}\Bigg(\dfrac{1024}{3}-\dfrac{1024}{5}-0\Bigg)\)
    \(=\dfrac{1}{64}\times\dfrac{2048}{15}\)
    \(=\dfrac{32}{15}\)

♦♦ Mean mark (b) 38%.
c.    \(\text{Pr}(2<T<4|T>1)\) \(=\dfrac{\text{Pr}(2<T<4)}{\text{Pr}(T>1)}\)
    \(=\dfrac{\dfrac{1}{64}\displaystyle \int_{2}^{4} (16t-t^3)\,dt}{\dfrac{1}{64}\displaystyle \int_{1}^{4} (16t-t^3)\,dt}\)
    \(=\dfrac{\left[8t^2-\dfrac{t^4}{4}\right]_2^4}{\left[8t^2-\dfrac{t^4}{4}\right]_1^4}\)
    \(=\dfrac{(64-(32-4))}{\bigg(64-\bigg(8-\dfrac{1}{4}\bigg)\bigg)}\)
    \(=\dfrac{36}{\bigg(\dfrac{225}{4}\bigg)}=\dfrac{144}{225}=\dfrac{16}{25}=0.64\)

♦♦♦ Mean mark (c) 24%.
MARKER’S COMMENT: Simplifying fractions caused problems. Cancelling factors will assist with calculations.

Networks, GEN2 2023 VCAA 14

One of the landmarks in state \(A\) requires a renovation project.

This project involves 12 activities, \(A\) to \(L\). The directed network below shows these activities and their completion times, in days.
 

The table below shows the 12 activities that need to be completed for the renovation project.

It also shows the earliest start time (EST), the duration, and the immediate predecessors for the activities.

The immediate predecessor(s) for activity \(I\) and the EST for activity \(J\) are missing.

\begin{array} {|c|c|c|}
\hline
\quad \textbf{Activity} \quad & \quad\quad\textbf{EST} \quad\quad& \quad\textbf{Duration}\quad & \textbf{Immediate} \\
&  & & \textbf{predecessor(s)}  \\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} & 0 & 6 & - \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 0 & 4 & - \\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 6 & 7 & A \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & 4 & 5 & B \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & 4 & 10 & B \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} & 13 & 4 & C \\
\hline
\rule{0pt}{2.5ex} G \rule[-1ex]{0pt}{0pt} & 9 & 3 & D \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} & 9 & 7 & D \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} & 13 & 6 & - \\
\hline
\rule{0pt}{2.5ex} J \rule[-1ex]{0pt}{0pt} & - & 6 & E, H \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} & 19 & 4 & F, I \\
\hline
\rule{0pt}{2.5ex} L \rule[-1ex]{0pt}{0pt} & 23 & 1 & J, K \\
\hline
\end{array}

  1. Write down the immediate predecessor(s) for activity \(I\).  (1 mark)

  2. What is the earliest start time, in days, for activity \(J\) ?  (1 mark)

  3. How many activities have a float time of zero?  (1 mark)

The managers of the project are able to reduce the time, in days, of six activities.

These reductions will result in an increase in the cost of completing the activity.

The maximum decrease in time of any activity is two days.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Activity} \rule[-1ex]{0pt}{0pt} & \quad \quad A \quad \quad & \quad \quad B \quad \quad&  \quad \quad F \quad \quad & \quad \quad H \quad \quad & \quad \quad I \quad \quad & \quad \quad K \quad \quad \\
\hline
\rule{0pt}{2.5ex} \textbf{Daily cost (\$)} \rule[-1ex]{0pt}{0pt} &  1500 & 2000 & 2500 & 1000 & 1500 & 3000  \\
\hline
\end{array}

  1. If activities \(A\) and \(B\) have their completion time reduced by two days each, the overall completion time of the project will be reduced.
  2. What will be the maximum reduction time, in days?  (1 mark)
  3. The managers of the project have a maximum budget of $15 000 to reduce the time for several activities to produce the maximum reduction in the project's overall completion time.
  4. Complete the table below, showing the reductions in individual activity completion times that would achieve the earliest completion time within the $ 15 000 budget.  (1 mark)

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} &  \\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} &  \\
\hline
\end{array}

Show Answers Only

a.    \(\text{Immediate predecessors of}\ I:\ C, G\)

b.    \(\text{EST}(J) = 4+5+7 = 16\ \text{days}\)

c.    \(\text{5 activities have a float time of zero.}\)

d.    \(\text{Maximum reduction time = 2 days}\)

e.    

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  0\\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} &  1\\
\hline
\end{array}

Show Worked Solution

a.    \(\text{Immediate predecessors of}\ I:\ C, G\)

\(\text{Dummy activity before activity}\ C\ \text{does not effect this.}\)
 

b.    \(\text{Scan network:}\)

\(\text{EST}(J) = 4+5+7 = 16\ \text{days}\)
 

c.    \(\text{Critical Path:}\ A\ C\ I\ K\ L\)

\(\text{Activities on the critical path have a float time of zero.}\)

\(\Rightarrow \ \text{5 activities have a float time of zero.}\)
 

d.    \(\text{If activities}\ A\ \text{and}\ B\ \text{are reduced by 2 days,}\)

\(\text{the critical path remains:}\ A\ C\ I\ K\ L\ \text{(22 days)}\)

\(\text{Maximum reduction time = 2 days}\)

♦♦ Mean mark (c) 35%.
♦♦ Mean mark (d) 33%.

e.    

\begin{array} {|c|c|}
\hline
\quad\textbf{Activity} \quad & \textbf{Reduction in completion time} \\
& \textbf{(0, 1 or 2 days)}\\
\hline
\rule{0pt}{2.5ex} A \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} F \rule[-1ex]{0pt}{0pt} &  0\\
\hline
\rule{0pt}{2.5ex} H \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} I \rule[-1ex]{0pt}{0pt} &  2\\
\hline
\rule{0pt}{2.5ex} K \rule[-1ex]{0pt}{0pt} &  1\\
\hline
\end{array}

♦♦♦ Mean mark (e) 9%.

Networks, GEN2 2023 VCAA 13

The state \(A\) has nine landmarks, \(G, H, I, J, K, L, M, N\) and \(O\).

The edges on the graph represent the roads between the landmarks.

The numbers on each edge represent the length, in kilometres, along each road.
 

 

Three friends, Eden, Reynold and Shyla, meet at landmark \(G\).

  1. Eden would like to visit landmark \(M\).
  2. What is the minimum distance Eden could travel from \(G\) to \(M\) ?  (1 mark)

  3. Reynold would like to visit all the landmarks and return to \(G\).
  4. Write down a route that Reynold could follow to minimise the total distance travelled.  (1 mark)

  5. Shyla would like to travel along all the roads.
  6. To complete this journey in the minimum distance, she will travel along two roads twice.
  7. Shyla will leave from landmark \(G\) but end at a different landmark.
  8. Complete the following by filling in the boxes provided.  
  9. The two roads that will be travelled along twice are the roads between:  (1 mark)

Show Answers Only

a.    \(\text{Shortest path:}\ G\ K\ I\ M\)

\(\text{Minimum distance}\ = 1.5+1.2+3.2 = 5.9\ \text{km}\)
 

b.   \(\text{Find the shortest Hamiltonian cycle starting at vertex}\ G:\)

\(G\ H\ K\ I\ J\ M\ O\ L\ N\ G\)

\(G\ N\ L\ O\ M\ J\ I\ K\ H\ G\ \text{(reverse of other path)}\)
 

c.   \(\text{vertex}\ L\ \text{and vertex}\ N\)

\(\text{vertex}\ J\ \text{and vertex}\ M\)

Show Worked Solution

a.    \(\text{Shortest path:}\ G\ K\ I\ M\)

\(\text{Minimum distance}\ = 1.5+1.2+3.2 = 5.9\ \text{km}\)
 

b.   \(\text{Find the shortest Hamiltonian cycle starting at vertex}\ G:\)

\(G\ H\ K\ I\ J\ M\ O\ L\ N\ G\)

\(G\ N\ L\ O\ M\ J\ I\ K\ H\ G\ \text{(reverse of other path)}\)

♦ Mean mark (b) 44%.

c.   \(\text{Using an educated guess and check methodology:}\)

\(\text{vertex}\ L\ \text{and vertex}\ N\)

\(\text{vertex}\ J\ \text{and vertex}\ M\)

♦♦♦ Mean mark (c) 12%.

Recursion and Finance, GEN2 2023 VCAA 7

Arthur takes out a new loan of $60 000 to pay for an overseas holiday.

Interest on this loan compounds weekly.

The balance of the loan, in dollars, after \(n\) weeks, \(V_n\), can be determined using a recurrence relation of the form

\(V_0=60\ 000, \quad V_{n+1}=1.0015\,V_n-d\)

  1. Show that the interest rate for this loan is 7.8% per annum.  (1 mark)

  2. Determine the value of \(d\) in the recurrence relation if
  3.   i. Arthur makes interest-only repayments  (1 mark)

  4.  ii. Arthur fully repays the loan in five years. Round your answer to the nearest cent.  (1 mark)

  5. Arthur decides that the value of \(d\) will be 300 for the first year of repayments.  
  6. If Arthur fully repays the loan with exactly three more years of repayments, what new value of \(d\) will apply for these three years? Round your answer to the nearest cent.  (1 mark)

  7. For what value of \(d\) does the recurrence relation generate a geometric sequence?  (1 mark)

Show Answers Only

 

Show Worked Solution

a.   \(\text{Weekly interest rate factor}\ = 1.0015-1 = 0.0015 = 0.15\% \)

\(I\%(\text{annual}) = 52 \times 0.15 = 7.8\%\)

♦♦ Mean mark (a) 32%.

 
b.i.  \(d= 0.15\% \times 60\ 000 = \dfrac{0.15}{100} \times 60\ 000 = 90\)
 

b.ii. \(\text{By TVM solver:} \)

\(N\) \(=5 \times 52 = 260\)  
\(I\%\) \(=7.8\)  
\(PV\) \(= -60\ 000\)  
\(PMT\) \(= ?\)  
\(FV\) \(=0\)  
\(\text{P/Y}\) \(=\ \text{C/Y}\ = 52\)  

 
\(\Rightarrow PMT = d= $278.86\)
 

♦ Mean mark (b)(i) 47%.
♦ Mean mark (b)(ii) 40%.

c.    \(d=300\ \text{for the 1st 52 weeks.}\)

\(\text{Find}\ FV\ \text{after 52 weeks:}\)

\(N\) \(=52\)  
\(I\%\) \(=7.8\)  
\(PV\) \(= -60\ 000\)  
\(PMT\) \(= 300\)  
\(FV\) \(=?\)  
\(\text{P/Y}\) \(=\ \text{C/Y}\ = 52\)  

 
\(\Rightarrow FV = $48\ 651.67\)

 
\(\text{Find}\ PMT\ \text{given}\ FV=0\ \text{after 156 more weeks:}\)

\(N\) \(=156\)  
\(I\%\) \(=7.8\)  
\(PV\) \(= -48\ 651.67\)  
\(PMT\) \(= ?\)  
\(FV\) \(=0\)  
\(\text{P/Y}\) \(=\ \text{C/Y}\ = 52\)  

 
\(\Rightarrow PMT = d= $350.01\)
 

♦♦ Mean mark (c) 32%.

d.    \(\text{Geometric sequence when}\ \ d=0\)

\(V_0=60\ 000, V_1=60\ 000(1.0015), V_2=60\ 000(1.0015)^2, …\)

♦♦♦ Mean mark (d) 11%.

Matrices, GEN1 2022 VCAA 8 MC

Two types of computers - laptops `(L)` and desktops `(D)` - can be serviced by Henry `(H)`, Irvine `(I)` or Jean `(J)`.

Matrix `N` shows the time, in minutes, it takes each person to service a laptop and a desktop.

`{:(qquadqquadquad\ LquadqquadD),(N = [(18,8),(10,17),(12,9)]{:(H),(I),(J):}):}`

Matrix `Q` shows the number of laptops and desktops in four different departments: marketing `(M)`, advertising `(A)`, publishing `(P)` and editing `(E)`.

`{:(qquadqquadquad\ LquadqquadD),(Q = [(6,8),(4,7),(5,5),(10,12)]{:(M),(A),(P),(E):}):}`

A calculation that determines the total time that it would take each of Henry, Irvine or Jean, working alone, to service all the laptops and desktops in all four departments is

  1. `[1\ 1\  1\  1]×(Q×N^T)`
  2. `(Q×N^T)×[(1),(1),(1)]`
  3. `(N×Q^T)×Q`
  4. `[(1,0,0),(0,1,0),(0,0,1)]×N×Q^T`
  5. `[1\ 1\ 1\ 1]×Q×N^T×[(1),(1),(1)]`
Show Answers Only

`A`

Show Worked Solution

`text{Required information requires a 3 × 1 or a 1 × 3 matrix.}`

`text{By elimination}`

`text{Option A: [1 × 4] × ([4 × 2][2 × 3]) → [1 × 3] }`

`text{Option B: ([4 × 2][2 × 3]) × [3 × 1] → [4 × 1] (eliminate B)}`

`text{Option C: ([3 × 2][2 × 4]) × [4 × 2] → [3 × 2] (eliminate C)}`

`text{Option D: [3 × 3][3 × 2][2 × 4] → [3 × 4] (eliminate D)}`

`text{Option E: [1 × 4][4 × 2][2 × 3][3 × 1] → [1 × 1] (eliminate E)}`

`=>A`

CHEMISTRY, M7 2020 VCE 16 MC

The following table provides information about three organic compounds, \(\text{X, Y}\) and \(\text{Z}\).

Which one of the following is the best estimate for the boiling point of Compound \(\text{Z}\)?

  1. 31 °C
  2. 101 °C
  3. 114 °C
  4. 156 °C
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\(A\)

Show Worked Solution

→ Boiling points of molecular compounds are related to the strength of their intermolecular bonding.

→ All given molecules are polar molecules. Each intermolecular bonding will have contributions from both dispersion forces and dipole-dipole bonding.

→ Since the molar mass of each compound is the same, the contribution from dispersion forces will be similar for all.

→ Compounds \(\text{X}\) and \(\text{Y}\) will both have hydrogen bonding, due to the presence of the \(\ce{O-H}\) functional group.

→ Compound \(\text{Z}\) does not have intermolecular hydrogen bonding and therefore would have the lowest boiling point.

\(\Rightarrow A\)

♦♦♦ Mean mark 20%.

Complex Numbers, EXT2 N2 2023 HSC 16c

The complex numbers \(w\) and \(z\) both have modulus 1, and  \(\dfrac{\pi}{2} \lt \text{Arg} \Big{(}\dfrac{z}{w}\Big{)} \lt \pi\), where \(\text{Arg}\) denotes the principal argument.

For real numbers \(x\) and \(y\), consider the complex number \(\dfrac{xz+yw}{z}\).

On an \(xy\)-plane, clearly sketch the region that contains all points \((x,y)\) for which \(\dfrac{\pi}{2} \lt \text{Arg} \Big{(}\dfrac{xz+yw}{z}\Big{)} \lt \pi\).

Show Answers Only

Show Worked Solution

\(\text{One strategy:}\)

\(\text{Let}\ \ u=\dfrac{z}{w},\ \text{where}\ \abs{z} = \abs{w} = 1\ \ \Rightarrow \ \ \abs{u}=1. \)

\(\dfrac{\pi}{2} \lt \text{Arg} (u) \lt \pi\ \text{(given)}\ \ \Rightarrow \text{Re}(u) \lt 0\ \ \text{and}\ \ \text{Im}(u) \gt 0 \)

\(\text{Using}\ \ \dfrac{z}{w} = \dfrac{1}{u} = \dfrac{\bar u}{\abs{u}} = \bar u, \)

\(\text{find all real numbers}\ x\ \text{and}\ y\ \text{such that:} \)

\(\dfrac{\pi}{2} \lt \text{Arg} \Big{(}\dfrac{xz+yw}{z}\Big{)} \lt \pi\)

\(\dfrac{\pi}{2} \lt \text{Arg} (x+ \bar y u) \lt \pi\)

\(\text{i.e.}\ \ \text{Re}(x+y \bar u) = x + y \times \text{Re}(u) \lt 0\ \ \text{and}\ \ \text{Im}(y \bar u) \gt 0 \)

  \(\therefore y \lt 0\ \ \text{and} \)

\(x \lt -y \times \text{Re}(u) = -y \times \cos (\text{Arg}(u)) = -y \times \cos \Bigg{(}\text{Arg}\Big{(} \dfrac{z}{w} \Big{)} \Bigg{)} \)

\(\text{Region is all real}\ (x,y)\ \text{that lie below the}\ x\text{-axis and to the} \)

\(\text{left of the line}\ \ y=-\sec \Bigg{(}\text{Arg}\Big{(} \dfrac{z}{w} \Big{)} \Bigg{)}x \)

♦♦♦ Mean mark 3%.
COMMENT: On the podium for the toughest questions in Ext2 in the last decade.

Vectors, EXT2 V1 2023 HSC 10 MC

Consider any three-dimensional vectors  \(\underset{\sim}{a}=\overrightarrow{O A}, \underset{\sim}{b}=\overrightarrow{O B}\)  and  \(\underset{\sim}{c}=\overrightarrow{O C}\)  that satisfy these three conditions

\(\underset{\sim}{a} \cdot \underset{\sim}{b}=1\)

\(\underset{\sim}{b} \cdot \underset{\sim}{c}=2\)

\(\underset{\sim}{c} \cdot \underset{\sim}{a}=3\).

Which of the following statements about the vectors is true?

  1. Two of \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) could be unit vectors.
  2. The points \(A, B\) and \(C\) could lie on a sphere centred at \(O\).
  3. For any three-dimensional vector \(\underset{\sim}{a}\), vectors \(\underset{\sim}{b}\) and \(\underset{\sim}{c}\) can be found so that \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) satisfy these three conditions.
  4. \(\forall \ \underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) satisfying the conditions, \(\exists \ r, s\) and \(t\) such that \(r, s\) and \(t\) are positive real numbers and  \(r\underset{\sim}{a}+s \underset{\sim}{b}+t \underset{\sim}{c}=\underset{\sim}{0}\).
Show Answers Only

\(B\)

Show Worked Solution

\(\text{By elimination}\)

\(\text{Option}\ A:\)

\(\text{If}\ \underset{\sim}{a}\ \text{and}\ \underset{\sim}{b}\ \text{are the two unit vectors,}\ \ \underset{\sim}{a} \cdot \underset{\sim}{b} = \abs{\underset{\sim}{a}} \abs{\underset{\sim}{b}} \cos\,\theta = \cos\,\theta\)

\(-1 \leq \cos\,\theta \leq1\ \ \Rightarrow \ \ -1 \leq \underset{\sim}{a} \cdot \underset{\sim}{b} \leq1 \)

\(\text{Given}\ \ \underset{\sim}{a} \cdot \underset{\sim}{b}=1\ \Rightarrow\ \underset{\sim}{a} = \underset{\sim}{b}\ \Rightarrow\ \underset{\sim}{b} \cdot \underset{\sim}{c} = \underset{\sim}{c} \cdot \underset{\sim}{a}\)

\(\text{Contradicts}\ \ \underset{\sim}{b} \cdot \underset{\sim}{c} = 2\ \ \text{and}\ \ \underset{\sim}{c} \cdot \underset{\sim}{a} = 3\)

\(\text{Similar reasoning rules out any pair satisfying all conditions (eliminate}\ A). \)
 

\(\text{Option}\ C:\ \text{If}\ \ \underset{\sim}{a} = \underset{\sim}{0}, \  \underset{\sim}{a} \cdot \underset{\sim}{b} = 0 \neq 1\ \text{(eliminate}\ C). \)

\(\text{Option}\ D:\)

\(\text{Consider the vectors below that satisfy the conditions,}\)

\[\underset{\sim}{a}=\left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right),\ \  \underset{\sim}{b}=\left(\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right), \ \  \underset{\sim}{c}=\left(\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right) \]

\(\text{However,}\ r\underset{\sim}{a}+s \underset{\sim}{b}+t \underset{\sim}{c}=\underset{\sim}{0}\ \text{requires}\ \ r=s=t=0\ \ \text{which are not}\)

\(\text{positive constants (eliminate}\ D).\)

\(\Rightarrow B\)

♦♦♦ Mean mark 15%.

Networks, GEN1 2023 VCAA 38 MC

A particular building project has ten activities that must be completed.

These activities and their immediate predecessor(s) are shown in the table below.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Activity} \rule[-1ex]{0pt}{0pt} & \textbf{Immediate predecessor(s)} \\
\hline
A  & - \\
\hline
B  & - \\
\hline
C  & A \\
\hline
 D  & A \\
\hline
E  & B \\
\hline
 F  & D, E \\
\hline
 G  & C, F \\
\hline
 H  & F \\
\hline
 I  & D, E \\
\hline
 J  & H, I \\
\hline
\end{array}

A directed graph that could represent this project is
 

Show Answers Only

\(D\)

Show Worked Solution

\(\text{Activity}\ H\ \text{has only one immediate predecessor,}\ F. \)

\(\text{Eliminate}\ A, B, C\ \text{and}\ E. \)

\(\Rightarrow D\)

Complex Numbers, EXT2 N2 2023 HSC 16a

Let \(w\) be the complex number  \(z=e^{\small{\dfrac{2i \pi}{3}}} \).

  1. Show that  \(1+w+w^2=0\).   (2 marks)

The vertices of a triangle can be labelled \(A, B\) and \(C\) in anticlockwise or clockwise direction, as shown.
 

Three complex numbers \(a, b\) and \(c\) are represented in the complex plane by points \(A, B\) and \(C\) respectively.

  1. Show that if triangle \(A B C\) is anticlockwise and equilateral, then  \(a+b w+c w^2=0\).   (2 marks)

  2. It can be shown that if triangle \(A B C\) is clockwise and equilateral, then \(a+b w^2+c w=0\). (Do NOT prove this.)
  3. Show that if \(A B C\) is an equilateral triangle, then

\(a^2+b^2+c^2=a b+b c+c a .\)  (2 marks)

Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(\text{See Worked Solutions}\)
  3. \(\text{See Worked Solutions}\)
Show Worked Solution

i.    \(w=e^{\small{\dfrac{2i \pi}{3}}} \)

\(w^2=e^{\small{-\dfrac{2i \pi}{3}}} = \overline w \)

\(\text{Re}(w)=\cos \Big{(}\dfrac{2\pi}{3} \Big{)}=-\dfrac{1}{2} \)

\(1+w+w^2\) \(=1+w+\overline w \)  
  \(=1+ 2 \times \text{Re}(w) \)  
  \(=1+2 \times -\dfrac{1}{2} \)  
  \(=0\)  

 
ii.    \(\text{Rotate}\ \ b-c\ \ \text{anticlockwise by}\ \ \dfrac{2\pi}{3} \)

\(\Rightarrow (b-c)w = c-a\)

\(\therefore bw-cw-c+a = a+bw-c(1+w) = 0 \)

\(\text{Substitute}\ \ 1+w=-w^2\ \ \text{(using part (i))} \)

\(a+bw-c(-w^2) \) \(=0\)  
\(a+bw+cw^2\) \(=0\ \ …\ \text{as required} \)  
♦♦♦ Mean mark (ii) 18%.

iii.   \(a+b w^2+c w=0\ \ \text{(if}\ \Delta ABC\ \text{is clockwise – given)} \)

\(a+bw+cw^2=0\ \ \text{(if}\ \Delta ABC\ \text{is anticlockwise)} \)

\(\Rightarrow (a+b w^2+c w)(a+bw+cw^2) = 0\ …\ (1)\)
 

\(\text{Expand (1) using part (i):}\ \ w+w^2=-1\ \ \text{and}\ \ w^3 = z^3 = 1 \)

\(a^2+abw^2+acw+bwa+b^2w^3+bcw^2+cw^2a+cbw^4+c^2w^3\) \(=0\)  
\(a^2+b^2+c^2+ab(w+w^2)+bc(w+w^2)+ca(w+w^2)\) \(=0\)  
\(a^2+b^2+c^2+ab(-1)+bc(-1)+ca(-1) \) \(=0\)  

 
\(\therefore a^2+b^2+c^2 = ab+bc+ca\ …\ \text{as required} \)

♦♦♦ Mean mark (iii) 11%.

Recursion and Finance, GEN1 2023 VCAA 23 MC

Tavi took out a loan of $20 000, with interest compounding quarterly. She makes quarterly repayments of $653.65.

The graph below represents the balance in dollars of Tavi's loan at the end of each quarter of the first year of the loan.
 

The effective interest rate for the first year of Tavi's loan is closest to

  1. 3.62%
  2. 3.65%
  3. 3.66%
  4. 3.67%
  5. 3.68%
Show Answers Only

\(D\)

Show Worked Solution

\(\text{After 1 quarter:}\)

\(\text{Principal paid}\ = 20\ 000-19\ 527.56= $472.44 \)

\(\text{Interest paid}\ = 653.65-472.44= $181.21\)

\(r = 4 \times \dfrac{181.21}{20\ 000} = 0.036242 = 3.6242\% \)

\(r_{effective}\) \(= \Bigg{[}\Big{(}1+\dfrac{r}{100n}\Big{)}^n-1 \Bigg{]} \times 100\%\)
  \(= \Bigg{[}\Big{(}1+\dfrac{0.03624}{4}\Big{)}^4-1 \Bigg{]} \times 100\%\)
  \(= 0.0367 \times 100\%\)
  \(= 3.67\%\)

 
\(\Rightarrow D\)

PHYSICS, M2 2023 VCE 8

Maia is at a skatepark. She stands on her skateboard as it rolls in a straight line down a gentle slope at a constant speed of 3.0 m s\(^{-1}\), as shown in Figure 8. The slope is 5° to the horizontal.

The combined mass of Maia and the skateboard is 65 kg.
  

  1. In Figure 9, the combined system of Maia and the skateboard is modelled as a small box with point \(\text{M}\) at the centre of mass.
  2. Draw and label arrows to represent each of the forces acting on the system - that is, Maia and skateboard as they roll down the slope.  (3 marks)
     


  1. Calculate the magnitude of the total frictional forces acting on Maia and the skateboard.  (2 marks)

Near the bottom of the ramp, Maia takes hold of a large pole and comes to a complete rest while still standing on the skateboard. Maia and the skateboard now have no momentum or kinetic energy.

  1. Explain what happened to both the momentum and the kinetic energy of Maia and the skateboard. No calculations are required.  (2 marks)

Show Answers Only

a.
       

b.     55.5 N

c.    Momentum:

→ Maia and the skateboards momentum was transferred into the pole and hence into the Earth. Due to the Earth’s very large mass, the effect on its velocity is negligible.

Kinetic energy:

→ The kinetic energy of Maia and the skateboard would have been transformed into heat energy between the contact of Maia and the pole and/or transferred into the work done by Maia’s muscles in slowing herself down.

Show Worked Solution

a.
       

b.     Total frictional forces:

→ Constant speed means that the force down the slope of the incline is equal to the sum of the frictional forces acting on Maia and the skateboard.

\(F_{\text{down slope}}\) \(=F_f\) \(=mg\, \sin\, \theta\)
    \(=65 \times 9.8 \times \sin\,5^{\circ}\)
    \(=55.5\ \text{N}\)

 

♦ Mean mark (b) 50%.

c.    Momentum:

→ Maia and the skateboards momentum was transferred into the pole and hence into the Earth. Due to the Earth’s very large mass, the effect on its velocity is negligible.

Kinetic energy:

→ The kinetic energy of Maia and the skateboard would have been transformed into heat energy between the contact of Maia and the pole and/or transferred into the work done by Maia’s muscles in slowing herself down.

♦♦♦ Mean mark (c) 26%.

ENGINEERING, PPT 2023 HSC 27b

A portion of a roller coaster wheel sub-assembly is shown.
 

An exploded pictorial of the wheel sub-assembly is shown.
 


 

Complete an assembled sectioned front view of the wheel sub-assembly at scale 1: 2. Apply AS 1100 drawing standards. Do NOT add dimensions.  (6 marks)

 

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Show Worked Solution


♦ Mean mark 49%.

ENGINEERING, CS 2023 HSC 26c

A truss is loaded as shown.
 


 

Showing working, complete the table.   (6 marks)
 

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt} & \textit{Magnitude} & \textit{Nature of force}\\ & \text{(kN)} & \text{(T or C)} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ EF \rule[-1ex]{0pt}{0pt} &  & \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ CH \rule[-1ex]{0pt}{0pt} &  & \\
\hline
\end{array}
Show Answers Only

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt} & \textit{Magnitude} & \textit{Nature of force}\\ & \text{(kN)} & \text{(T or C)} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ EF \rule[-1ex]{0pt}{0pt} & 73.2\ \text{kN} & \text{Compression} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ CH \rule[-1ex]{0pt}{0pt} & 71.138\ \text{kN} & \text{Tension} \\
\hline
\end{array}

Show Worked Solution

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt} & \textit{Magnitude} & \textit{Nature of force}\\ & \text{(kN)} & \text{(T or C)} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ EF \rule[-1ex]{0pt}{0pt} & 73.2\ \text{kN} & \text{Compression} \\
\hline
\rule{0pt}{2.5ex} \text{Internal reaction of member}\ CH \rule[-1ex]{0pt}{0pt} & 71.138\ \text{kN} & \text{Tension} \\
\hline
\end{array}

\(\text{Consider member}\ EF:\)

               

→ \(\text{Force diagram closes with two collinear forces}\)

→ \(FG\ \text{is a zero force member}\)

→ \(EF = 73.2\ \text{kN (compression)} \)

♦♦♦ Mean mark 36%.

\(\text{Consider member}\ CH:\)
 

\(+ \uparrow \Sigma F_{V}\) \(=0\)  
\(0\) \(=-125 + 73.2 \times \sin\,60^{\circ} + CH \times \sin\,60^{\circ}\)  
\(CH \times \sin\,60^{\circ}\) \(=61.607\)  
\(CH\) \(= \dfrac{61.607}{\sin\,60^{\circ}} \)  
  \(=71.138\ \text{kN (tension)} \)  

ENGINEERING, PPT 2023 HSC 25d

A uniform 8-metre ladder with a mass of 12 kg has been placed against a smooth wall. 
 

Determine the minimum coefficient of static friction between the ground and the ladder. Assume there is no friction between the ladder and the wall.   (4 marks)

 

Show Answers Only

\(\mu = 0.287\)

Show Worked Solution

Graphical solution:

\(\phi = 16^{\circ}\)

\(\mu = \tan \phi = \tan 16^{\circ} = 0.287\ \text{(3 d.p.)}\)

♦♦ Mean mark 31%.

Analytical solution:

\(\cos 60^{\circ} = \dfrac{\text{d}_1}{4}\ \Rightarrow \ \text{d}_1 = 2\ \text{m} \)

\(\sin 60^{\circ} = \dfrac{\text{d}_2}{8}\ \Rightarrow \ \text{d}_2 = 6.928\ \text{m} \)

\( \Sigma \text{M}_{\text{G}} \) \(=0\)  
\(0\) \(=(120 \times 2)-(\text{F}_{\text{W}} \times 6.928) \)  
\(\text{F}_{\text{W}}\) \(=\dfrac{240}{6.928}\)  
  \(= 34.64\ \text{N}\)  

 
\(\therefore \text{F}_{\text{F}} = 34.64 \text{ N}\)
 

\(\text{F}_{\text{g}}\) \(\ =mg\)  
  \(\ =120\ \text{N}\)  

 
\(\therefore \text{N} = 120 \text{N}\uparrow \)
 

\(\text{F}_{\text{F}}\) \(= \mu \text{N}\)  
\(\mu\) \(= \dfrac{34.64}{120} \)  
  \(=0.289\ \text{(3 d.p.)} \)  

ENGINEERING, PPT 2023 HSC 24b

A roller coaster component is fabricated using cold rolled steel. Two webs are welded onto a base plate as shown.   (2 marks)
 

The diagram shows a partially completed microstructure of the parent and weld metals.

Complete the microstructure by drawing and labelling the following grain types in the heat-affected zone for ONE of the webs:

    • chill crystals
    • equi-axed grains
    • columnar grains.

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Correct answers include one of the labelled grain types:

Show Worked Solution

Correct answers include one of the labelled grain types:
 

BIOLOGY, M2 EQ-Bank 29

The countercurrent flow in the gills of fish allow for up to 95% of oxygen to be extracted from water.

Explain how the structure of fish gills and the countercurrent flow contribute to the efficiency of the gas exchange process.   (4 marks)

Show Answers Only

→ Fish do not breathe air like humans but instead acquire oxygen through water when it flows over their gills.

→ The gill is structured by a gill bar (in bony fish) and contains thousands of leaf like filaments called lamellae.

→ Gas exchange can occur when water flows over the lamellae and their abundance drastically increases surface area, allowing for a higher frequency of gas exchange.

→ Gills also utilise countercurrent flow, where the water and deoxygenated blood will flow in opposite directions.

→ This allows for a concentration gradient to be maintained, rather than co-current flow which results in diffusion stopping when the oxygen is ‘evenly split’ 50/50 between the water and the blood.

Show Worked Solution

→ Fish do not breathe air like humans but instead acquire oxygen through water when it flows over their gills.

→ The gill is structured by a gill bar (in bony fish) and contains thousands of leaf like filaments called lamellae.

→ Gas exchange can occur when water flows over the lamellae and their abundance drastically increases surface area, allowing for a higher frequency of gas exchange.

→ Gills also utilise countercurrent flow, where the water and deoxygenated blood will flow in opposite directions.

→ This allows for a concentration gradient to be maintained, rather than co-current flow which results in diffusion stopping when the oxygen is ‘evenly split’ 50/50 between the water and the blood.

PHYSICS, M3 EQ-Bank 5

A 50 gram copper ball is placed into an insulated container containing 50 mL of water and immediately sealed. The initial temperature of the metal ball and water is 50°C and 10°C respectively. 

A student hypothesises that since the water and copper ball both have the same mass, the temperature of the metal ball and water, once thermal equilibrium is established, would be 30°C.

When the student measured the temperature inside the container, it was 26°C.

Explain the results of the experiment and why the student's hypothesis is incorrect.   (4 marks)

Show Answers Only

→ Although the masses of the copper ball and water are the same, the specific heat capacities of the objects are different which leads to different changes in temperature.

→ The student’s hypothesis is incorrect as they did not take the specific heat capacity values into account.

→ The specific heat capacity of water is greater than that of copper, thus a greater amount of energy would be required to heat water to a certain temperature than to heat copper to that same temperature.

→ Therefore, when reaching a state of thermal equilibrium, the energy transfer between the copper ball and water cools the copper ball down faster than the water heats up.

→ This leads to the final temperature within the container of 26°C, which is closer to the initial temperature of the water than the copper ball.

Show Worked Solution

→ Although the masses of the copper ball and water are the same, the specific heat capacities of the objects are different which leads to different changes in temperature.

→ The student’s hypothesis is incorrect as they did not take the specific heat capacity values into account.

→ The specific heat capacity of water is greater than that of copper, thus a greater amount of energy would be required to heat water to a certain temperature than to heat copper to that same temperature.

→ Therefore, when reaching a state of thermal equilibrium, the energy transfer between the copper ball and water cools the copper ball down faster than the water heats up.

→ This leads to the final temperature within the container of 26°C, which is closer to the initial temperature of the water than the copper ball.

PHYSICS, M3 EQ-Bank 1

Jack enjoys drinking green tea at 60°C.

He pours 200 mL of 90°C green tea into a cup and then adds 4°C chilled water to the same cup to cool it down.

What is the minimum amount of chilled water, to the nearest mL, required to cool Jack's green tea to 60°C?   (4 marks)

(Assume the green tea is essentially water and no energy is lost to the surroundings) 

Show Answers Only

\(107\ \text{mL}\)

Show Worked Solution

→ The energy contained in the temperature of the green tea that is lost to reduce it to 60°C needs to be used to heat up the chilled water from 4°C to 60°C.

→ Green tea’s change in temperature \(=90^{\circ}-60^{\circ}=30^{\circ}\).

→ The energy that needs to be lost from the green tea can be quantised:

\(Q=mc\Delta t=200 \times 4.18 \times -30=-25\ 080\ \text{J}\)

→ This \(25\ 080\ \text{J}\) then goes into heating up a required amount of chilled water from \(4^{\circ}\) to \(60^{\circ}\).

\(Q\) \(=mc\Delta t\)  
\(25\ 080\) \(=m \times 4.18 \times 56\)  
\(m\) \(=\dfrac{25\ 080}{4.18 \times 56}\)  
  \(=107\ \text{g}\)  
  \(=107\ \text{mL}\)  

PHYSICS, M1 EQ-Bank 5

Car A approaches an intersection at 10 ms\(^{-1}\) from the South as shown. As Car B approaches the intersection, it measures the velocity of Car A to be 22.36 ms\(^{-1}\) NW.
 

Draw Car B on the diagram showing its direction and speed. Show all working.  (3 marks)

Show Answers Only

→ Velocity of A relative to B = \(v_A-v_B=v_A+-v_B\)
 

→ The direction of \(-v_B\) must be west, and therefore the direction of \(v_B\) is east.

→ Using Pythagoras theorem, the magnitude of \(-v_B\) is:

\(-v_B\) \(=\sqrt{22.36^2-10^2}\)  
  \(=20\ \text{ms}^{-1}\)  

 

→ Thus the vector \(v_B\) is \(20\ \text{ms}^{-1}\) east.
 

Show Worked Solution

→ Velocity of A relative to B = \(v_A-v_B=v_A+-v_B\)
  

→ The direction of \(-v_B\) must be west, and therefore the direction of \(v_B\) is east.

→ Using Pythagoras theorem, the magnitude of \(-v_B\) is:

\(-v_B\) \(=\sqrt{22.36^2-10^2}\)  
  \(=20\ \text{ms}^{-1}\)  

 

→ Thus the vector \(v_B\) is \(20\ \text{ms}^{-1}\) east.
 

PHYSICS, M3 2018 VCE 11

Figure 13 shows two speakers, \(\text{A}\) and \(\text{ B}\), facing each other. The speakers are connected to the same signal generator/amplifier and the speakers are simultaneously producing the same 340 Hz sound.
 

Take the speed of sound to be 340 ms\( ^{-1}\).

  1. Calculate the wavelength of the sound.  (1 mark)

  2. A student stands in the centre, equidistant from speakers \(\text{A}\) and \(\text{B}\). He then moves towards speaker \(\text{B}\) and experiences a sequence of loud and quiet regions. He stops at the second region of quietness.
  3. How far has the student moved from the centre? Explain your reasoning.  (3 marks)

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a.     \(\lambda=1\ \text{m}\)

b.    → At the centre of the speakers, there is an area of relative loudness

→ This is due to a path difference of 0 metres between the wavelengths of sounds coming from both speaker \(\text{A}\) and speaker \(\text{B}\). 

→ As the student moves towards speaker \(\text{B}\), the two sound waves will experience constructive (loud) and destructive interference (quiet).

→ The peak of speaker \(\text{A}\) will move into the trough of speaker \(\text{B}\) where he experiences the first region of quietness (node) which is 1 quarter wavelength (0.25m) from the centre.

→ The second region of quietness (node) will be half of a wavelength (0.5m) from the first region of quietness. Thus, the total distance from the centre to the second region of quietness is 0.75m.

Show Worked Solution

a.    \(\lambda=\dfrac{v}{f}=\dfrac{340}{340}=1\ \text{m}\)

 

b.    → At the centre of the speakers, there is an area of relative loudness

→ This is due to a path difference of 0 metres between the wavelengths of sounds coming from both speaker \(\text{A}\) and speaker \(\text{B}\). 

→ As the student moves towards speaker \(\text{B}\), the two sound waves will experience constructive and destructive interference.

→ The peak of speaker \(\text{A}\) will move into the trough of speaker \(\text{B}\) where he experiences the first region of quietness (node) which is 1 quarter wavelength (0.25m) from the centre.

→ The second region of quietness (node) will be half of a wavelength (0.5m) from the first region of quietness. Thus, the total distance from the centre to the second region of quietness is 0.75m.

♦♦♦ Mean mark (b) 20%.

CHEMISTRY, M2 2014 VCE 11*

Consider the following unbalanced ionic equation.

\(\ce{Hg(l) + Cr2O7^2–(aq) + H+(aq)\rightarrow Hg^2+(aq) + Cr^3+(aq) + H2O(l)}\)

When this equation is completely balanced including the total charge on each side of the equation, find the coefficient of \(\ce{Hg(l)}\).   (3 marks)

Show Answers Only

\(3\)

Show Worked Solution

→ By balancing the equations by the total number of atoms of each side

\(\ce{Hg(l) + Cr2O7^2–(aq) + 14H+(aq)\rightarrow Hg^2+(aq) + 2Cr^3+(aq) + 7H2O(l)}\)

→ The charge on the left side is \(+12\) while the charge on the right side is \(+8\)

→ The charge on the right hand side needs to be increased by \(+4\)

→ This can be done by changing to \(\ce{3Hg^2+(aq)}\).

→ The total balanced equation while following law of conservation of charge is:

\(\ce{3Hg(l) + Cr2O7^2–(aq) + 14H+(aq)\rightarrow 3Hg^2+(aq) + 2Cr^3+(aq) + 7H2O(l)}\)

→ Hence the coefficient of \(\ce{Hg(l)}\) is \(3\).

PHYSICS, M4 2020 VCE 18

Students are modelling the effect of the resistance of electrical cables, \(r\), on the transmission of electrical power. They model the cables using the circuit shown in Figure 18.
 

The students investigate the effect of changing \(r\) by measuring the current in the electrical cables for a range of values. Their results are shown in Table 1 below.
 

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Resistance of cables,}\ r\ (\Omega) \rule[-1ex]{0pt}{0pt} & \text{Current in cables},\ i\ (\text{A}) & \ \ \ \dfrac{1}{i} \Big{(}\text{A}^{-1}\Big{)}\ \ \  \\
\hline
\rule{0pt}{2.5ex} 2.4 \rule[-1ex]{0pt}{0pt} & 2.4 &  \\
\hline
\rule{0pt}{2.5ex} 3.6 \rule[-1ex]{0pt}{0pt} & 2.0 &  \\
\hline
\rule{0pt}{2.5ex} 6.4 \rule[-1ex]{0pt}{0pt} & 1.7 &  \\
\hline
\rule{0pt}{2.5ex} 7.6 \rule[-1ex]{0pt}{0pt} & 1.5 &  \\
\hline
\rule{0pt}{2.5ex} 10.4 \rule[-1ex]{0pt}{0pt} & 1.3 &  \\
\hline
\end{array}

  1. To analyse the data, the students use the following equation to calculate the resistance of the cables for the circuit.
  2.         \(r=\dfrac{24}{i}-R\)
  3. Show that this equation is true for the circuit shown in Figure 18. Show your working.  (2 marks)

  4. Calculate the values of \(\dfrac{1}{i}\) and write them in the spaces provided in the last column of Table 1 . (1 mark)

  5. Plot a graph of \(r\) on the \(y\)-axis against \(\dfrac{1}{i}\) on the \(x\)-axis on the grid provided below. On your graph:
    • choose an appropriate scale and numbers for the \(x\)-axis
    • draw a straight line of best fit through the plotted points  (3 marks)


  1. Use the straight line of best fit to find the value of the constant resistance globe, \(R\). Give your reasoning.  (2 marks)

Show Answers Only

a.      \(V\) \(=iR\)  
\(V\) \(=i(r+R)\)  
\(\dfrac{V}{i}\) \(=r+R\)  
\(r\) \(=\dfrac{V}{i}-R\)  
\(r\) \(=\dfrac{24}{i}-R\)  

b.   

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Resistance of cables,}\ r\ (\Omega) \rule[-1ex]{0pt}{0pt} & \text{Current in cables},\ i\ (\text{A}) & \ \ \ \dfrac{1}{i} \Big{(}\text{A}^{-1}\Big{)}\ \ \  \\
\hline
\rule{0pt}{2.5ex} 2.4 \rule[-1ex]{0pt}{0pt} & 2.4 & 0.42 \\
\hline
\rule{0pt}{2.5ex} 3.6 \rule[-1ex]{0pt}{0pt} & 2.0 & 0.5 \\
\hline
\rule{0pt}{2.5ex} 6.4 \rule[-1ex]{0pt}{0pt} & 1.7 & 0.59 \\
\hline
\rule{0pt}{2.5ex} 7.6 \rule[-1ex]{0pt}{0pt} & 1.5 & 0.67 \\
\hline
\rule{0pt}{2.5ex} 10.4 \rule[-1ex]{0pt}{0pt} & 1.3 & 0.77 \\
\hline
\end{array}

c.    
       

d.    → The equation for the line:  \(r=\dfrac{24}{i}-R\).

→ The \(y\)-intercept of the graph correlates to the value of \(R\).

→ Reading from the graph, \(R= 7\ \Omega\).

Show Worked Solution

a.      \(V\) \(=iR\)  
\(V\) \(=i(r+R)\)  
\(\dfrac{V}{i}\) \(=r+R\)  
\(r\) \(=\dfrac{V}{i}-R\)  
\(r\) \(=\dfrac{24}{i}-R\)  
♦♦ Mean mark (a) 37%.

b.   

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Resistance of cables,}\ r\ (\Omega) \rule[-1ex]{0pt}{0pt} & \text{Current in cables},\ i\ (\text{A}) & \ \ \ \dfrac{1}{i} \Big{(}\text{A}^{-1}\Big{)}\ \ \  \\
\hline
\rule{0pt}{2.5ex} 2.4 \rule[-1ex]{0pt}{0pt} & 2.4 & 0.42 \\
\hline
\rule{0pt}{2.5ex} 3.6 \rule[-1ex]{0pt}{0pt} & 2.0 & 0.5 \\
\hline
\rule{0pt}{2.5ex} 6.4 \rule[-1ex]{0pt}{0pt} & 1.7 & 0.59 \\
\hline
\rule{0pt}{2.5ex} 7.6 \rule[-1ex]{0pt}{0pt} & 1.5 & 0.67 \\
\hline
\rule{0pt}{2.5ex} 10.4 \rule[-1ex]{0pt}{0pt} & 1.3 & 0.77 \\
\hline
\end{array}

c.    
       

d.    → The equation for the line:  \(r=\dfrac{24}{i}-R\).

→ The \(y\)-intercept of the graph correlates to the value of \(R\).

→ Reading from the graph, \(R= 7\ \Omega\).

♦♦ Mean mark (d) 33%.

PHYSICS, M3 2021 VCE 14

A distant fire truck travelling at 20 ms\(^{-1}\) to a fire has its siren emitting sound at a constant frequency of 500 Hz.

Chris is standing on the edge of the road. Assume that the fire truck is travelling directly towards him as it approaches and directly away from him as it goes past. The arrangement is shown in Figure 14.
 

  1. On the diagram below, sketch the frequency that Chris will hear as the truck moves towards him and then moves away from him. The \(500 \text{ Hz}\) siren signal is shown as a dotted line for reference. No calculations are required.  (2 marks)
     


 

  1. Name the physics principle involved in Chris’s experience.  (1 mark)
Show Answers Only

a.    

  
b.   The Doppler effect
Show Worked Solution

a.    

♦♦♦ Mean mark (a) 13%.
COMMENT: Poorly completed by most students, many incorrectly drew a smooth line that passed Chris at 500 Hz.
  
b.   The Doppler effect

PHYSICS, M2 2021 VCE 4

Liesel, a student of yoga, sits on the floor in the lotus pose, as shown in Figure 4. The action force, \(F_g\), on Liesel due to gravity is 500 N down.
 

Identify and explain what the reaction force is to the action force, \(F_{ g }\), shown in Figure 4.  (2 marks)

Show Answers Only

→ As the action force, \(F_g\), is the gravitational force on Liesel due to the Earth, then the reaction force is the gravitational force on the Earth due to Liesel, by Newton’s third law of motion.

→ This would be a force of 500 N of Liesel pulling up on the Earth (The Earth being attracted\accelerated to Liesel due to her own gravitational force).

Show Worked Solution

→ As the action force, \(F_g\), is the gravitational force on Liesel due to the Earth, then the reaction force is the gravitational force on the Earth due to Liesel, by Newton’s third law of motion.

→ This would be a force of 500 N of Liesel pulling up on the Earth (the Earth being attracted\accelerated to Liesel due to her own gravitational force).

♦♦♦ Mean mark 9%.
COMMENT: Many students confused Newton’s third law with balancing forces. The normal force of 500N pushing up on Liesel is the normal force not the reaction force (which are not the same).

PHYSICS, M4 2022 VCE 4*

Two point charges, \(Q\) and \(4Q\), are placed 12 cm apart, as shown in the diagram below.
 

On the straight line between the charges \(Q\) and \(4Q\), find where the electric field is zero.   (3 marks)

Show Answers Only

\(\text{4cm to the right of}\ Q\)

Show Worked Solution

→ The point where the force on a charged particle \(q\) as a result of the two fields from \(Q\) and \(4Q\) is equal and opposite is where the electric field strength would be zero.

→ Let \(x\) equal the distance from \(Q\) to \(q\) and \(y\) be the distance from \(4Q\) to \(q\) so  \(x +y=12\)
 

\(F_{\text{\(Q\) on \(q\)}}\) \(=F_{\text{\(4Q\) on \(q\)}}\)  
\(\dfrac{1}{4 \pi \varepsilon_0} \dfrac{Q \times q}{x^2}\) \(=\dfrac{1}{4 \pi \varepsilon_0} \dfrac{4Q \times q}{y^2}\)  
\(\dfrac{1}{x^2}\) \(=\dfrac{4}{y^2}\)  
\(y^2\) \(=4x^2\)  
\(y\) \(=2x\)  

 

→ \(x+2x=12\ \ \Rightarrow\ \ x=4\ \text{cm} \)

♦♦♦ Mean mark 18%.

PHYSICS, M1 2012 HSC 21

  1. Outline a first-hand investigation that could be performed to measure a value for acceleration due to gravity.   (3 marks)

  2. How would you assess the accuracy of the result of the investigation?   (1 mark)

  3. How would you increase the reliability of the data collected?   (1 mark)

  4. How would you assess the reliability of the data collected?   (1 mark)

Show Only

a.    Timing of a falling mass.

→ Set up an electronic and automatic timing system with sensors to detect the presence of a small falling metal ball.

→ Heights for the ball should be set up between 0.2 m to 1 m with intervals every 0.2 m. To increase the reliability of the results, multiple trials should be conducted at each height and the average falling time for each height should be calculated which can then be used to graph the data.

→ The results should be plotted on a graph of height vs time\(^2\). This uses the equation  \(s=ut +\dfrac{1}{2}at^2\)  where  \(u=0\)  which becomes  \(s=\dfrac{1}{2}at^2\).

→ After plotting the data, the acceleration due to gravity, \(a\), can be calculated using  \(a=\dfrac{2s}{t^2}\), which will make it equal to 2 × the gradient of the line of best fit.
 

b.    → Look up known value on a reliable website (e.g. National Measurement Institute).

→ Ensure the value is for the location of the experiment (it can differ slightly).

→ Compare the known value to the value determined experimentally and the closer they are, the greater the accuracy of the experiment.
 

c.    → Conduct multiple trials at each height.

→ Use the average of the calculations as stated in the method above.
 

d.   → Compare the values obtained at a single height.

→ If there is a large variation in the calculations conducted at the same height, the data collected is less reliable.

Show Worked Solution

a.    Timing of a falling mass.

→ Set up an electronic and automatic timing system with sensors to detect the presence of a small falling metal ball.

→ Heights for the ball should be set up between 0.2 m to 1 m with intervals every 0.2 m. To increase the reliability of the results, multiple trials should be conducted at each height and the average falling time for each height should be calculated which can then be used to graph the data.

→ The results should be plotted on a graph of height vs time\(^2\). This uses the equation  \(s=ut +\dfrac{1}{2}at^2\)  where  \(u=0\)  which becomes  \(s=\dfrac{1}{2}at^2\).

→ After plotting the data, the acceleration due to gravity, \(a\), can be calculated using  \(a=\dfrac{2s}{t^2}\), which will make it equal to 2 × the gradient of the line of best fit.

♦ Mean mark (a) 55%.

b.    → Look up known value on a reliable website (e.g. National Measurement Institute).

→ Ensure the value is for the location of the experiment (it can differ slightly).

→ Compare the known value to the value determined experimentally and the closer they are, the greater the accuracy of the experiment.

♦♦♦ Mean mark (b) 26%.

c.    → Conduct multiple trials at each height.

→ Use the average of the calculations as stated in the method above.

 
d.   → Compare the values obtained at a single height.

→ If there is a large variation in the calculations conducted at the same height, the data collected is less reliable.

♦♦♦ Mean mark (d) 7%.

Complex Numbers, EXT2 N1 2023 HSC 7 MC

Which of the following statements about complex numbers is true?

  1. For all real numbers \(x, y, \theta\)  with  \(x \neq 0\),

\(\tan \theta=\dfrac{y}{x} \ \Rightarrow \ x+i y=r e^{i \theta}\), for some real number \(r\).

  1. For all non-zero complex numbers \(z_1\) and \(z_2\),

\(\operatorname{Arg}\left(z_1\right)=\theta_1\)  and  \(\operatorname{Arg}\left(z_2\right)=\theta_2 \ \Rightarrow \ \operatorname{Arg}\left(z_1 z_2\right)=\theta_1+\theta_2,\)

where \(\operatorname{Arg}\) denotes the principal argument.

  1. For all real numbers \(r_1, r_2, \theta_1, \theta_2\)  with  \(r_1, r_2>0\),

\(r_1 e^{i \theta_1}=r_2 e^{i \theta_2} \ \Rightarrow \ r_1=r_2\)  and  \(\theta_1=\theta_2 \text {. }\)

  1. For all real numbers \(x, y, r, \theta\)  with  \(r>0\)  and  \(x \neq 0\),

\(x+i y=r e^{i \theta} \ \Rightarrow \ \theta=\arctan  \Big(\dfrac{y}{x} \Big)\)

Show Answers Only

\(A\)

Show Worked Solution

\(\text{Eliminating options by contradiction}\)

\(\text{Option}\ B:\)

\(\text{If}\ \ \theta_1= \pi\ \ \text{and}\ \ \theta_2=\dfrac{\pi}{2}, \ \theta_1 + \theta_2 = \dfrac{3\pi}{2} > \pi \)

\( -\pi < \operatorname{Arg}\left(z_1 z_2\right) < \pi\ \ \ \  \text{(Eliminate}\ B) \)

♦♦♦ Mean mark 18%.

\(\text{Option}\ C:\)

\(\text{If}\ \ \theta_1= \pi\ \ \text{and}\ \ \theta_2=3\pi, \ \operatorname{Arg}(e^{i\pi}) =  \operatorname{Arg}(e^{3i\pi}) \)

\( \text{However,}\ \ \theta_1 \neq \theta_2\ \ \ \  \text{(Eliminate}\ C) \)
 

\(\text{Option}\ D:\)

\(\text{If}\ \ x=y=-1, \ \theta=-\dfrac{3\pi}{4} \ \ (r>0) \)

\( \text{However,}\ \ \arctan\Big(\dfrac{-1}{-1}\Big)=\dfrac{\pi}{4} \ \ \  \text{(Eliminate}\ D) \)

\(\Rightarrow A\)

PHYSICS, M4 2017 HSC 8 MC

An electron is fired in a vacuum towards a screen. With no electric field being applied, the electron hits the screen at \(P\). A uniform electric field is turned on and another electron is fired towards the screen from the same location, at the same velocity, striking the screen at point \(Q\).

With the electric field still turned on, a proton is fired towards the screen from the same starting point as the electrons and with the same velocity.

At what point does the proton strike the screen?

  1. \(A\)
  2. \(B\)
  3. \(C\)
  4. \(D\)
Show Answers Only

\(C\)

Show Worked Solution

→ As the proton has an equal but opposite charge to the electron it will experience a force in the opposite direction. Hence it will hit the screen to the right of \(P\).

→ Due to the proton having a greater mass, it will be deflected by a smaller amount and so will hit the screen at \(C\).

\(\Rightarrow C\)

♦♦♦ Mean mark 25%.

CHEMISTRY, M8 2023 HSC 36

An organic reaction pathway involving compounds \(\text{A, B,}\) and \(\text{C}\) is shown in the flow chart.
 

The molar mass of \(\text{A}\) is 84.156 g mol\(^{-1}\).

A chemist obtained some spectral data for the compounds as shown.
 

\( \text{Data from} \ ^{1} \text{H NMR spectrum of compound C} \)
 \( Chemical \ Shift \ \text{(ppm)} \) \( Relative \ peak \ area \) \( Splitting \ pattern \)
\(1.01\) \(3\) \(\text{Triplet}\)
\(1.05\) \(3\) \(\text{Triplet}\)
\(1.65\) \(2\) \(\text{Multiplet}\)
\(2.42\) \(2\) \(\text{Triplet}\)
\(2.46\) \(2\) \(\text{Quartet}\)
\( ^{1} \text{H NMR chemical shift data}\)
\( Type \ of \ proton \)  \(  \text{δ/ppm} \)
\( \ce{R - C\textbf{H}3,R - C\textbf{H}2 - R}\) \(0.7-1.7\)
\( \left.\begin{array}{l}\ce{\textbf{H}3C - CO - \\-C\textbf{H}2 - CO -}\end{array}\right\} \begin{aligned} & \text { (aldehydes, ketones,} \\ &\text{carboxylic acids or esters) }\end{aligned}\) \(2.0-2.6\) 
\( \ce{R - C\textbf{H}O} \) \(9.4-10.00\)
\( \ce{R - COO\textbf{H}} \) \(9.0-13.0\)
 

Identify the functional group present in each of compounds \(\text{A}\) to \(\text{C}\) and draw the structure of each compound. Justify your answer with reference to the information provided.  (9 marks) 

Show Answers Only

Compound \(\text{A}\): Alkene
 

Compound \(\text{B}\): Secondary alcohol
 

Compound \(\text{C}\): Ketone
 

Reasoning as follows:

→ Compound \(\text{A}\) is able to undergo an addition reaction to add water across a \(\ce{C=C}\) bond  \(\Rightarrow \)  Alkene

→ Compound \(\text{B}\) is the product of the above hydration reaction and is therefore an alcohol.

→ The \(\ce{^{13}C\ NMR}\) spectrum of Compound \(\text{A}\) confirms it is an alkene (132 ppm peak corresponding to the \(\ce{C=C}\) atoms). 3 spectrum peaks indicate 3 carbon environments. The molar mass of compound \(\text{A}\) is 84.156 g mol\(^{-1}\) which suggests symmetry within the molecule.

→ The Infrared Spectrum of Compound \(\text{B}\) has a broad peak at approximately 3400 cm\(^{-1}\). This indicates the presence of an hydroxyl group and confirms \(\text{B}\) is an alcohol.

→ Compound \(\text{C}\) is produced by the oxidation of Compound \(\text{B}\) with acidified potassium permanganate.

→ Compound \(\text{C}\) is a carboxylic acid if \(\text{B}\) is a primary alcohol or a ketone if \(\text{B}\) is a secondary alcohol.

→ Since the \(\ce{^{1}H NMR}\) spectrum of \(\text{C}\) does not show any peaks between 9.0 − 13.0 ppm, it cannot be a carboxylic acid. Compound \(\text{C}\) is therefore a ketone and Compound \(\text{B}\) is a secondary alcohol.

→ The \(\ce{^{1}H NMR}\) spectrum shows 5 peaks \(\Rightarrow \) 5 hydrogen environments.

→ Chemical shift and splitting patterns information indicate:

1.01 ppm – 1.05 ppm: \(\ce{CH3}\) (next to a \(\ce{CH2}\))

1.65 ppm: \(\ce{CH2}\) (with multiple neighbouring hydrogens)

2.42 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH2}\))

2.46 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH3}\))

Show Worked Solution

Compound \(\text{A}\): Alkene
 

Compound \(\text{B}\): Secondary alcohol
 

Compound \(\text{C}\): Ketone
 

Reasoning as follows:

→ Compound \(\text{A}\) is able to undergo an addition reaction to add water across a \(\ce{C=C}\) bond  \(\Rightarrow \)  Alkene

→ Compound \(\text{B}\) is the product of the above hydration reaction and is therefore an alcohol.

→ The \(\ce{^{13}C\ NMR}\) spectrum of Compound \(\text{A}\) confirms it is an alkene (132 ppm peak corresponding to the \(\ce{C=C}\) atoms). 3 spectrum peaks indicate 3 carbon environments. The molar mass of compound \(\text{A}\) is 84.156 g mol\(^{-1}\) which suggests symmetry within the molecule.

→ The Infrared Spectrum of Compound \(\text{B}\) has a broad peak at approximately 3400 cm\(^{-1}\). This indicates the presence of an hydroxyl group and confirms \(\text{B}\) is an alcohol.

→ Compound \(\text{C}\) is produced by the oxidation of Compound \(\text{B}\) with acidified potassium permanganate.

→ Compound \(\text{C}\) is a carboxylic acid if \(\text{B}\) is a primary alcohol or a ketone if \(\text{B}\) is a secondary alcohol.

→ Since the \(\ce{^{1}H NMR}\) spectrum of \(\text{C}\) does not show any peaks between 9.0 − 13.0 ppm, it cannot be a carboxylic acid. Compound \(\text{C}\) is therefore a ketone and Compound \(\text{B}\) is a secondary alcohol.

→ The \(\ce{^{1}H NMR}\) spectrum shows 5 peaks \(\Rightarrow \) 5 hydrogen environments.

→ Chemical shift and splitting patterns information indicate:

1.01 ppm – 1.05 ppm: \(\ce{CH3}\) (next to a \(\ce{CH2}\))

1.65 ppm: \(\ce{CH2}\) (with multiple neighbouring hydrogens)

2.42 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH2}\))

2.46 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH3}\))