Band 3 – Page 5

PHYSICS, M6 2019 VCE 5-6* MC

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

Question 5

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

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

Question 6

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

$0.04\ \text{A}$
$0.20\ \text{A}$
$1.20\ \text{A}$
$1.50\ \text{A}$

Show Answers Only

$\text{Question 5:}\ B$

$\text{Question 6:}\ C$

Show Worked Solution

$\text{Question 5}$

→ The primary coil is connected to the power supply (i.e. Part $\text{X}$).

→ For a step-up transformer, there are more coils in the secondary coil then there are in the primary coil.

$\Rightarrow B$

$\text{Question 6}$

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

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

$\Rightarrow C$

Mean mark Q6 57%.

Calculus, MET1 2023 VCE SM-Bank 6

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

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

--- 4 WORK AREA LINES (style=lined) ---
Using an initial estimate of $x_0=-1$, find the value of $x_1$. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $\text{See worked solutions}$

b. $-\dfrac{14}{9}$

Show Worked Solution

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

b.	$x_1$	$=x_0-\dfrac{f(x_0)}{f^{\prime}(x_0)},\quad\ \text{where }x_0=-1$
		$=-1-\dfrac{-\dfrac{1}{3}+2.(-1)+4}{(-1)^2+2}$
		$=-\dfrac{14}{9}$

Functions, MET1 2023 VCE SM-Bank 2

Consider the functions $f$ and $g$, where

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

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

--- 2 WORK AREA LINES (style=lined) ---
Determine the rule for the equation and state the domain of the function $f \circ g$. (2 marks)

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

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $[-9, \infty)$

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

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

Show Worked Solution

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

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

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

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

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

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

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

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

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

Graphs, MET1 2023 VCE SM-Bank 1

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

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

--- 0 WORK AREA LINES (style=lined) ---

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

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

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

Show Worked Solution

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

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

Calculus, MET2 2023 VCE SM-Bank 2

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

The cross-section of the ramp is modelled by the function $f$, where

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

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

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

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

--- 2 WORK AREA LINES (style=lined) ---
i. Find the coordinates of the point of inflection of $f$. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
ii. Find the interval of $x$ for which the gradient function of the ramp is strictly increasing. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
iii. Find the interval of $x$ for which the gradient function of the ramp is strictly decreasing. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---

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

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

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

--- 4 WORK AREA LINES (style=lined) ---
Jac and Jill roll the toy car down the ramp and the car jumps off the end of the ramp. The path of the car is modelled by the function $P$, where
1. 1. $P(x)=\begin{cases}f(x) & 0 \leq x \leq 55 \\ g(x) & 55<x \leq a\end{cases}$
$P$ is continuous and differentiable at $x=55, g(x)=-\frac{1}{16} x^2+b x+c$, and $x=a$ is where the car lands on the ground after the jump, such that $P(a)=0$.

1. Find the values of $b$ and $c$. (2 marks)
  
  --- 6 WORK AREA LINES (style=lined) ---
2. Determine the horizontal distance from the end of the ramp to where the car lands. Give your answer in centimetres, correct to two decimal places. (1 mark)
  
  --- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

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

b.i $(25, 20)$

b.ii $[25, 55]$

b.iii $[5, 25]$

c. $98.1\%$

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

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

$\text{making the trapezium rule approximation greater than the}$

$\text{actual area.}$

e.i $b=8.75, c=-283.4375$

e.ii $34.10\ \text{cm}$

Show Worked Solution

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

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

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

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

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

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

c. $\text{Using CAS:}$

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

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

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

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

$\text{making the trapezium rule approximation greater than the}$

$\text{actual area.}$

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

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

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

$g'(x)=-\dfrac{x}{8}+b$

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

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

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

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

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

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

$c=-\dfrac{4535}{16}$

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

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

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

PHYSICS, M7 2020 VCE 16 MC

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

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

sodium, potassium, lithium, nickel
nickel, potassium, sodium, lithium
potassium, nickel, lithium, sodium
lithium, sodium, potassium, nickel

Show Answers Only

$D$

Show Worked Solution

→ $K_{\text{max}}=hf-\Phi \ \ \Rightarrow\ \ \Phi=hf-K_{\text{max}}$

→ The threshold frequency occurs when $K_{\text{max}}=0.$

→ $\Phi=hf_{\text{thresh}}$, hence $\Phi \propto f_{\text{thresh}}$.

→ The greater the threshold frequency, the greater the work function. Therefore, lithium has the greatest work function.

$\Rightarrow D$

Statistics, SPEC2 2022 VCAA 6

A company produces soft drinks in aluminium cans.

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

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

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

Write down suitable hypotheses $H_0$ and $H_1$ for this test. (1 mark)
Find the $p$ value for the test, correct to three decimal places. (1 mark)
Does the mean mass of the random sample of 64 empty cans support the supplier's claim at the 5% level of significance for a one-tailed test? Justify your answer. (1 mark)
What is the smallest value of the mean mass of the sample of 64 empty cans for $H_0$ not to be rejected? Give your answer correct to two decimal places. (1 mark)

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

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

Show Answers Only

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

b. $p=0.027$

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

d. $a=14.95$

e. $\text{Pr}(-3<D<3)=0.329$

Show Worked Solution

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

b. $\mu=15, \ \ \sigma=0.25$

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

$\text{By CAS:}$

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

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

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

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

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

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

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

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

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

$\text {Let}\ \ D=M_1-M_2$

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

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

$\sigma_D=\sqrt{50}$

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

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

♦ Mean mark (e) 46%.

CHEMISTRY, M2 EQ-Bank 8

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

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

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

a. Determine the empirical formula of each compound.

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

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

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

→ Butyraldehyde and lactic acid have the same empirical formula of $\ce{CH2O}$.

b. $\ce{C16H20O8}$

Show Worked Solution

a. Determine the empirical formula of each compound.

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

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

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

→ Butyraldehyde and lactic acid have the same empirical formula of $\ce{CH2O}$.

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

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

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

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

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

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

→ Thus, the molecular formula of the compound is $\ce{C16H20O8}$.

CHEMISTRY, M2 EQ-Bank 4

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

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

$226.8\ \text{mol}$

Show Worked Solution

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

$\ce{n(C3H8) = \dfrac{\text{m}}{\text{MM}} = \dfrac{10\,000\ \text{g}}{44.094 \, \text{g mol}^{-1}} = 226.79 \, mol}$

→ The number of moles of propane in the tank is 226.8 mol.

Calculus, SPEC2 2022 VCAA 3

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

i. Express the differential equation in the form $\displaystyle \int g(x)dx=\int f(t)dt$. (1 mark)
ii. Hence, show that $x=\log _e\left(\tan ^{-1}(2 t)+1\right)$. (2 marks)
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)

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

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

Show Answers Only

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

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

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

b.ii.

c. $0.02$

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

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

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

e. $\dfrac{v_1}{v_2}=\dfrac{2}{3}$

Show Worked Solution

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

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

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

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

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

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

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

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

♦♦♦ Mean mark 23%.

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

$\text{When}\ \ t=10 :$

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

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

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

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

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

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

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

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

$\text {At}\ \ t=6:$

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

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

$\therefore \dfrac{v_1}{v_2}=\dfrac{2}{3}$

CHEMISTRY, M1 EQ-Bank 29

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

Suggest a suitable separation technique to extract the salt from the saltwater solution. Explain your reasoning based on the physical property involved. (2 marks)
--- 4 WORK AREA LINES (style=blank) ---
Identify the physical property that allow the mixture of sand and iron fillings to be separated and whether it is a homogeneous or heterogeneous mixture. (1 marks)
--- 2 WORK AREA LINES (style=blank) ---
Discuss one safety precaution that should be followed during the separation of the saltwater solution. (2 marks)
--- 4 WORK AREA LINES (style=blank) ---

Show Answers Only

a. Separation technique:

→ Evaporation is a suitable technique because water has a much lower boiling point than salt.

→ By heating the solution, the water evaporates, leaving salt crystals behind.

→ Further distillation could be used to collect and condense the evaporated water.

b. Physical property:

→ Iron fillings are magnetic and hence can be separated using a magnet.

→ The mixture is heterogeneous.

c. Safety precaution:

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

Show Worked Solution

a. Separation technique:

→ Evaporation is a suitable technique because water has a much lower boiling point than salt.

→ By heating the solution, the water evaporates, leaving salt crystals behind.

→ Further distillation could be used to collect and condense the evaporated water.

b. Physical property:

→ Iron fillings are magnetic and hence can be separated using a magnet.

→ The mixture is heterogeneous.

c. Safety precaution:

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

Functions, MET2 2022 VCAA 4

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

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

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

--- 1 WORK AREA LINES (style=lined) ---
i. Find `f^{\prime}(0)`. (2 marks)

--- 3 WORK AREA LINES (style=lined) ---
ii. State the maximal domain over which `f` is strictly increasing. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Show that `f(x)+f(-x)=0`. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Find the domain and the rule of `f^{-1}`, the inverse of `f`. (3 marks)

--- 4 WORK AREA LINES (style=lined) ---
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`.
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)}`.
The area of the regions bound by the functions `h` and `h^{-1}` can be expressed as a function, `A(k)`.
The graph below shows the relevant area shaded.

You are not required to find or define `A(k)`.

Determine the range of values of `k` such that `A(k)>0`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Explain why the domain of `A(k)` does not include all values of `k`. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

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

Show Worked Solution

a. `R` is the range.

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

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

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

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

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

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

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

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

`\frac{1}{k}(4)<1` [Using CAS]

`:.\ k > 4`

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

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

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

♦♦♦♦ Mean mark (e.ii) 10%.

Calculus, SPEC2 2022 VCAA 1

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

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

--- 3 WORK AREA LINES (style=lined) ---
Sketch the graph of $y=f(x)$ for $k=1$ on the set of axes below. Clearly label any turning points with their coordinates and label any asymptotes with their equations. (3 marks)

--- 0 WORK AREA LINES (style=lined) ---

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

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

--- 4 WORK AREA LINES (style=lined) ---
Now consider the functions $h$ and $g$, where $h(x)=x+3$ and $g(x)=\abs{\dfrac{x^2}{x-1}}$.
The region bounded by the curves of $h$ and $g$ is rotated about the $x$-axis.

1. Write down the definite integral that can be used to find the volume of the resulting solid. (2 marks)
  
  --- 5 WORK AREA LINES (style=lined) ---
2. Hence, find the volume of this solid. Give your answer correct to two decimal places. (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

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

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

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

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

d.ii. $V=51.42\ \text{u}^3 $

Show Worked Solution

a. $\text {When } k=1 :$

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

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

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

$\text {Using part a.}$

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

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

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

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

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

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

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

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

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

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

♦♦ Mean mark (d)(ii) 37%.

Vectors, SPEC2 2022 VCAA 13 MC

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

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

$-\cos (t) \underset{\sim}{\text{i}}+2 \sin (t) \underset{\sim}{\text{j}}$
$(3-\cos (t)) \underset{\sim}{\text{i}}+(2 \sin (t)+1) \underset{\sim}{\text{j}}$
$(1+\cos (t)) \underset{\sim}{\text{i}}+(2\sin (t)+1) \underset{\sim}{\text{j}}$
$(2+\sin (t)) \underset{\sim}{\text{i}}+(2\cos (t)-1) \underset{\sim}{\text{j}}$
$(1+\cos (t)) \underset{\sim}{\text{i}}+(1-2\sin (t)) \underset{\sim}{\text{j}}$

Show Answers Only

$B$

Show Worked Solution

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

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

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

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

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

$=>B$

PHYSICS, M6 2020 VCE 6*

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

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

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

$8.8 \times 10^{-5}\ \text{V}$

Show Worked Solution

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

PHYSICS, M5 2020 VCE 2*

Jupiter's moon Ganymede is its largest satellite.

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

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

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

$1.5\ \text{ms}^{-2}$

Show Worked Solution

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

Functions, EXT1 F1 2023 MET1 7

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

State the range of $f$. (1 mark)
--- 2 WORK AREA LINES (style=lined) ---
Sketch the graph of the inverse function $y=f^{-1}(x)$ on the axes above. Label any endpoints and axial intercepts with their coordinates. (2 marks)
--- 0 WORK AREA LINES (style=lined) ---
Determine the equation of the domain for the inverse function $f^{-1}$. (2 marks)
--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $[-1, \infty)$

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

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

Show Worked Solution

a. $[-1, \infty)$

c. $\text{When }f(x)\ \text{is written in turning point form}$

$y=(x-1)^2-1$

$\text{Inverse function: swap}\ x \leftrightarrow y$

$x$	$=(y-1)^2-1$
$x+1$	$=(y-1)^2$
$-\sqrt{x+1}$	$=y-1$
$f^{-1}(x)$	$=1-\sqrt{x+1}$

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

♦ Mean mark (c) 48%.
MARKER’S COMMENT: Common error → writing the function as $f^{-1}(x)=1+\sqrt{x+1}$.

Functions, MET2 2022 VCAA 15 MC

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

`(-\infty, \infty)`
`(-\infty,-3) \cup(1, \infty)`
`(-1,3)`
`[-3,1]`
`(-\infty,-1] \cup[3, \infty)`

Show Answers Only

`E`

Show Worked Solution

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

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

`:. \ (x – 3)(x + 1)>=0`

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

`=>E`

Probability, MET2 2022 VCAA 10 MC

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

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

(4.1, 6.9)
(50.9, 59.1)
(52.4, 57.6)
(51.9, 58.1)
(45.2, 64.8)

Show Answers Only

`D`

Show Worked Solution

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

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

`=>D`

Calculus, MET2 2022 VCAA 8 MC

If `\int_0^b f(x)dx=10` and `\int_0^a f(x)dx=-4`, where `0<a<b`, then `\int_a^b f(x)dx` is equal to

-6
-4
0
10
14

Show Answers Only

`E`

Show Worked Solution

`\int_0^b f(x)dx`	`=\int_0^a f(x)dx + \int_a^b f(x)dx`
`10`	`= -4 + \int_a^b f(x)dx`
`:.\ \int_a^b f(x)dx`	`= 10 + 4 = 14`

`=>E`

Graphs, MET2 2022 VCAA 2 MC

The graph of `y=\frac{1}{(x+3)^2}+4` has a horizontal asymptote with the equation

`y=4`
`y=3`
`y=0`
`x=-2`
`x=-3`

Show Answers Only

`A`

Show Worked Solution

From graph

`=>A`

Graphs, SPEC2 2022 VCAA 1 MC

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

`y=-x-2`
`y=3x-4`
`y=x+2`
`y=3x+2`
`y=x-4`

Show Answers Only

`B`

Show Worked Solution

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

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

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

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

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

`=>B`

Vectors, EXT2 V1 2022 SPEC1 6

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

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`8/21`

Show Worked Solution

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

Vectors, SPEC1 2022 VCAA 6a

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

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

`8/21`

Show Worked Solution

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

Complex Numbers, EXT2 N2 2022 SPEC1 1

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

Express `p(z)` in the form `p(z) = (z+ai)^2 + b` where `a`, ` b ∈ R`. (1 mark)
Hence, or otherwise, find the solutions of the equation `p(z) = 0`. (2 marks)

Show Answers Only

`p(z) = (z + 3i)^2-16`
`z = ± 4-3i`

Show Worked Solution

a.	`p(z)`	`= (z + 3i)^2-(3i)^2-25`
		`=(z + 3i)^2 +9-25`
		`=(z + 3i)^2-16`

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

Complex Numbers, SPEC1 2022 VCAA 1

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

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

--- 3 WORK AREA LINES (style=lined) ---
Hence, or otherwise, find the solutions of the equation `p(z) = 0`. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

`p(z) = (z + 3i)^2-16`
`z = ± 4-3i`

Show Worked Solution

a.	`p(z)`	`= (z + 3i)^2-(3i)^2-25`
		`=(z + 3i)^2 +9-25`
		`=(z + 3i)^2-16`

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

Statistics, SPEC2 2023 VCAA 6

A forest ranger wishes to investigate the mass of adult male koalas in a Victorian forest. A random sample of 20 such koalas has a sample mean of 11.39 kg.

It is known that the mass of adult male koalas in the forest is normally distributed with a standard deviation of 1 kg.

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

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

--- 2 WORK AREA LINES (style=lined) ---

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

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

--- 2 WORK AREA LINES (style=lined) ---

It is thought that the mean mass of adult male koalas in the forest is 12 kg. The ranger thinks that the true mean mass is less than this and decides to apply a one-tailed statistical test. A random sample of 40 adult male koalas is taken and the sample mean is found to be 11.6 kg.

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

--- 2 WORK AREA LINES (style=lined) ---

The ranger decides to apply the one-tailed test at the 1% level of significance and assumes the mass of adult male koalas in the forest is normally distributed with a mean of 12 kg and a standard deviation of 1 kg.

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

--- 2 WORK AREA LINES (style=lined) ---
ii. Draw a conclusion about the null hypothesis in part d. from the $p$ value found above, giving a reason for your conclusion. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
What is the critical sample mean (the smallest sample mean for $H_0$ not to be rejected) in this test? Give your answer in kilograms correct to three decimal places. (1 mark)
--- 2 WORK AREA LINES (style=lined) ---

Suppose that the true mean mass of adult male koalas in the forest is 11.4 kg, and the standard deviation is 1 kg. The level of significance of the test is still 1%.

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

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $(10.95,11.83)$

b. $57$

c. $n=125$

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

e.i. $p=0.0057$

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

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

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

Show Worked Solution

a. $\sigma_{\text{pop}}=1$

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

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

$(10.95,11.83)$

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

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

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

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

$\text{Interval reduced by } 60\%$

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

$\text{Solve for } n$ :

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

$\Rightarrow n=125$

♦♦♦ Mean mark (c) 28%.

d. $H_0: \mu=12,\ \ H_1: \mu<12$

e.i. $E(\bar{X})=\mu=12$

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

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

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

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

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

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

$\text {Critical sample mean}\ \ \bar{x} \approx 11.632$

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

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

♦♦ Mean mark (g) 39%.

PHYSICS, M7 2022 VCE 14*

Sam undertakes a photoelectric effect experiment using the apparatus shown in Figure 12. She uses a green filter.

Sam produces a graph of photocurrent, $I$, in milliamperes, versus voltage, $V$, in volts, as shown in Figure 13.

Identify what point $\text{P}$ represents on the graph in Figure 13. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Sam then significantly increases the intensity of the light.
Sketch the resulting graph on Figure 14. The dashed line in Figure 14 represents the original data. (2 marks)

--- 0 WORK AREA LINES (style=blank) ---

Sam replaces the green filter with a violet filter, keeping the light source at the increased intensity.
Sketch the resulting graph on Figure 15. The dashed line in Figure 15 represents the original data. (2 marks)

--- 0 WORK AREA LINES (style=blank) ---

Show Answers Only

a. Stopping voltage

Show Worked Solution

a. Stopping voltage

b. Diagram will exhibit the following points:

→ Increasing the intensity of light will increase the photocurrent due to more photoelectrons being ejected from the metal.

→ However, it will not change the energy of the photoelectrons, hence the stopping voltage remains the same.

c. Diagram will exhibit the following points:

→ Changing the colour from green to violet will increase the frequency of the light.

→ As $E=hf$, the energy of the photoelectrons will increase, hence there needs to be a greater stopping voltage.

Calculus, MET2 2023 VCAA 1

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

State the coordinates of all axial intercepts of $f$. (1 mark)
--- 2 WORK AREA LINES (style=lined) ---
Find the coordinates of the stationary points of $f$. (2 marks)
--- 3 WORK AREA LINES (style=lined) ---
1. Let $g:R\rightarrow R, g(x)=x-2$.
2. Find the values of $x$ for which $f(x)=g(x)$. (1 mark)
  
  --- 3 WORK AREA LINES (style=lined) ---

1. Write down an expression using definite integrals that gives the area of the regions bound by $f$ and $g$. (2 marks)
  --- 4 WORK AREA LINES (style=lined) ---
2. Hence, find the total area of the regions bound by $f$ and $g$, correct to two decimal places. (1 mark)
  --- 2 WORK AREA LINES (style=lined) ---

Let $h:R\rightarrow R, h(x)=(x-a)(x-b)^2$, where $h(x)=f(x)+k$ and $a, b, k \in R$.
Find the possible values of $a$ and $b$. (4 marks)
--- 9 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $(-1, 0), (0, 0), (2, 0)$

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

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

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

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

c.iii. $5.95$

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

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

Show Worked Solution

a. $(-1, 0), (0, 0), (2, 0)$

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

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

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

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

$\text{The stationary points of }f\ \text{are}:$

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

ci $\text{Given }f(x)=g(x)$

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

$\therefore\ \text{Using CAS: } $

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

cii $\text{Area of bounded region:}$

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

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

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

d. $\text{Method 1 – Equating coefficients}$

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

$x^3-2bx^2-ax^2+b^2x+2abx-ab^2=x^3-x^2-2x+k$

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

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

$2ab+b^2=-2\ \ …(2)$

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

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

$\text{Method 2 – Using transformations}$

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

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

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

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

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

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

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

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

PHYSICS, M5 2021 VCE 9-10 MC

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

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

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

Question 9

Which one of the following expressions correctly gives the distance $d$ ?

0.8$t$
6$t$
5$t^2$
6$t$ + 5$t^2$

Question 10

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

0.8 s
1.1 s
1.3 s
1.6 s

Show Answers Only

$\text{Question 9:} \ B$

$\text{Question 10:} \ C$

Show Worked Solution

$\text{Question 9}$

→ The horizontal displacement formula for projectile motion: $s=ut$

→ Horizontal distance: $d=6t$

$\Rightarrow B$

$\text{Question 10}$

$\text{Find time of flight}\ (t):$

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

$\Rightarrow C$

Functions, 2ADV F2 2023 MET1 3

Sketch the graph of $f(x)=2-\dfrac{3}{x-1}$ on the axes below, labelling all asymptotes with their equation and axial intercepts with their coordinates. (3 marks)

--- 0 WORK AREA LINES (style=lined) ---
Find the values of $x$ for which $f(x)\leq1$. (1 mark)
--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

b. $1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]$

Show Worked Solution

a. $\text{Vertical asymptote when}\ \ x=1$

$y\text{-int:}\ y=2-(-3)\ \ \Rightarrow\ \ y=5$

$x\text{-int:}\ 2-\dfrac{3}{x-1}=0\ \ \Rightarrow\ \ x=\dfrac{5}{2} $

$\text{As}\ \ x \rightarrow \infty, \ \ y \rightarrow 2^{-}; \ \ x \rightarrow -\infty, \ \ y \rightarrow 2^{+} $

b. $\text{From the graph:}$

$f(x)=1\ \text{when }x=4$

$x>1\ \text{to the right of the vertical asymptote}$

$\therefore\ f(x)\leq1\ \text{when}\ \ 1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]$

♦♦ Mean mark (b) 38%.
MARKER’S COMMENT: Many students did not use their graph from part (a).

PHYSICS, M5 2022 VCE 8

A Formula 1 racing car is travelling at a constant speed of 144 km h$^{-1}$ (40 m s$^{-1}$) around a horizontal corner of radius 80.0 m. The combined mass of the driver and the car is 800 kg. Figure 8a shows a front view and Figure 8b shows a top view.

Calculate the magnitude of the net force acting on the racing car and driver as they go around the corner. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

On Figure 8b, draw the direction of the net force acting on the racing car using an arrow. (1 mark)

--- 0 WORK AREA LINES (style=blank) ---

Explain why the racing car needs a net horizontal force to travel around the corner and state what exerts this horizontal force. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $1.6 \times 10^4\ \text{N}$

c. → A net horizontal force is required for circular motion.

→ The horizontal force is perpendicular to the direction of travel.

→ This is the force which changes the direction of travel.

→ The force of friction from the road acts on the tires provides the centripetal force in this motion.

Show Worked Solution

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

c. → A net horizontal force is required for circular motion.

→ The horizontal force is perpendicular to the direction of travel.

→ This is the force which changes the direction of travel.

→ The force of friction from the road acts on the tires provides the centripetal force in this motion.

♦ Mean mark 42%.

PHYSICS, M6 2022 VCE 6

Figure 6 shows a simple alternator consisting of a rectangular coil of area 0.060 m$^{2}$ and 200 turns, rotating in a uniform magnetic field. The magnetic flux through the coil in the vertical position shown in Figure 6 is 1.2 × 10$^{-3}$ Wb.

Calculate the strength of the magnetic field in Figure 6 . Show your working. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

The rectangular coil rotates at a frequency of 2.5 Hz.
Calculate the average induced EMF produced in the first quarter of a turn. Begin the quarter with the coil in the vertical position shown in Figure 6. (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $0.02\ \text{T}$

b. $2.4\ \text{V}$

Show Worked Solution

a.	$\Phi$	$=BA$
	$B$	$=\dfrac{\Phi}{A}$
		$=\dfrac{1.2 \times 10^{-3}}{0.060}$
		$=0.02\ \text{T}$

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

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

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

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

PHYSICS, M6 2022 VCE 5*

A wind generator provides power to a factory located 2.00 km away, as shown in Figure 5.

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

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

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

--- 2 WORK AREA LINES (style=lined) ---

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

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

--- 6 WORK AREA LINES (style=lined) ---

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

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

--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $41.5\ \text{kW}$

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

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

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

c. $41.3\ \text{kW}$

Show Worked Solution

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

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

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

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

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

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

$\Rightarrow\ \text{Net power supplied}\ =41.5-0.2=41.3\ \text{kW}$

PHYSICS, M6 2022 VCE 1

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

When the motor is operating, the coil rotates about the axis through the middle of sides $L M$ and $N K$ in the direction indicated. The coil is attached to a commutator. Current for the motor is passed to the commutator by brushes that are not shown in Figure 1.

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

--- 3 WORK AREA LINES (style=lined) ---

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

--- 2 WORK AREA LINES (style=lined) ---

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

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

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $K$ to $L$

b. $L$ to $K$

c. $3\ \text{A}$

Show Worked Solution

a. Using the right hand rule:

→ Palm faces up (force), fingers to the right (magnetic field), and the thumb faces out of the page.

→ Therefore the current must run from $K$ to $L$.

b. Using the right hand rule:

→ Palm now faces down the page (force), fingers still to the right.

→ The current will run from $L$ to $K$ (thumb).

c.	$F$	$=lIB$
	$I$	$=\dfrac{F}{lB}$
		$=\dfrac{0.15}{0.1 \times 0.5}$
		$=3\ \text{A}$

PHYSICS, M7 2022 VCE 18 MC

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

a bus travelling at constant velocity
an express train that is accelerating
a car turning a corner at a constant speed
a roller-coaster speeding up while heading down a slope

Show Answers Only

$A$

Show Worked Solution

→ $A$ is the only frame of reference where no acceleration is imposed on the object.

→ It is therefore an inertial frame of reference.

$\Rightarrow A$

PHYSICS, M7 2022 VCE 16 MC

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

polarisation
interference
dispersion
diffraction

Show Answers Only

$A$

Show Worked Solution

→ Polarisation requires the transfer of energy through a wave to be perpendicular to the direction of oscillation of the wave.

→ As this is a property of a transverse wave, light waves must be traverse.

$\Rightarrow A$

PHYSICS, M7 2022 VCE 15 MC

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

photoelectric effect
white light passing through a prism
diffraction of light through a single slit
interference of light passing through a double slit

Show Answers Only

$A$

Show Worked Solution

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

$\Rightarrow A$

PHYSICS, M8 2022 VCE 3 MC

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

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

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

	$ \textbf{K} $	$ \textbf{L} $	$ \textbf{M} $
A.	positive	no charge	negative
B.	positive	negative	negative
C.	negative	no charge	positive
D.	no charge	no charge	no charge

Show Answers Only

$A$

Show Worked Solution

→ $\text{L}$ must have no charge as it travels through the magnetic field with no deflection.

→ Using the right-hand rule, the force on a positive charge would be up the page, hence $\text{K}$ must be the positive charge.

$\Rightarrow A$

PHYSICS, M6 2022 VCE 1 MC

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

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

to the left
to the right
into the page
out of the page

Show Answers Only

$C$

Show Worked Solution

→ Right hand grip rule: fingers curl in a clockwise direction and the thumb points into the page.

$\Rightarrow C$

Calculus, SPEC2 2023 VCAA 4

A fish farmer releases 200 fish into a pond that originally contained no fish. The fish population, $P$, grows according to the logistic model, $\dfrac{d P}{d t}=P\left(1-\dfrac{P}{1000}\right)$ , where $t$ is the time in years after the release of the 200 fish.

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

--- 2 WORK AREA LINES (style=lined) ---

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

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

--- 2 WORK AREA LINES (style=lined) ---

The farmer releases a batch of $n$ fish into a second pond, pond 2 , which originally contained no fish. The population, $Q$, of fish in pond 2 can be modelled by $Q=\dfrac{1000}{1+9 e^{-1.1 t}}$, where $t$ is the time in years after the $n$ fish are released.

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

--- 2 WORK AREA LINES (style=lined) ---
Find the value of $Q$ when $t=6$.
Give your answer correct to the nearest integer. (1 mark)

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

--- 4 WORK AREA LINES (style=lined) ---
ii. Hence or otherwise, find the size of the fish population in pond 2 and the value of $t$ when the rate of growth of the population is a maximum. Give your answer for $t$ correct to the nearest year. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Sketch the graph of $Q$ versus $t$ on the set of axes below. Label any axis intercepts and any asymptotes with their equations. (1 mark)
--- 0 WORK AREA LINES (style=lined) ---

The farmer wishes to take 5.5% of the fish from pond 2 each year. The modified logistic differential equation that would model the fish population, $Q$, in pond 2 after $t$ years in this situation is

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

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

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $A=1, \quad B=\dfrac{1}{1000}$

b. $\Rightarrow D=4$

c. $n=\dfrac{1000}{1+9 e^0}=100$

d. $Q=\dfrac{1000}{1+9 e^{-6.6}} \approx 988$

e.i. $Q^{\prime}=\frac{121}{100}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{121}{100\ 000}\left(1-\frac{Q}{1000}\right)$

e.ii. $t=2$

g. $Q=950$

Show Worked Solution

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

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

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

$A=1, \quad B=\dfrac{1}{1000}$

♦ Mean mark (a) 48%.

b. $\text{When}\ \ t=0, P=200 \text{ (given)}$

$\text{Solve}\ \ 200=\dfrac{1000}{1+D e^0}\ \ \text{for}\ t:$

$\Rightarrow D=4\ \ \text {(by CAS):}$

c. $Q=\dfrac{1000}{1+9 e^{-1.1 t}}$

$\text{At}\ \ t=0, Q=n$:

$n=\dfrac{1000}{1+9 e^0}=100$

d. $\text{Find } Q \text{ when } t=6:$

$Q=\dfrac{1000}{1+9 e^{-6.6}} \approx 988$

e.i. $\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)$

$\text{Using the product rule:}$

\begin{aligned}
Q^{\prime \prime} & =\frac{11}{10}\left[Q^{\prime}\left(1-\frac{Q}{1000}\right)+Q\left(-\frac{1}{1000}\, Q^{\prime}\right)\right] \\
& =\frac{11}{10}\left[\frac{11}{10}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{Q}{1000}\left(\frac{11}{10}\, Q\left(1-\frac{Q}{1000}\right)\right)\right] \\
& =\frac{121}{100}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{121}{100\ 000}\left(1-\frac{Q}{1000}\right)
\end{aligned}

♦♦♦ Mean mark (e)(i) 21%.

e.ii. $\text{Max } Q^{\prime} \Rightarrow Q^{\prime \prime}=0$

$\text{Solve } Q^{\prime \prime}=0 \ \ \text {(by CAS):}$

$\Rightarrow Q=500$

$\text{Solve } Q=500 \text{ for } t \text{ (by CAS):}$

$t=1.99 \ldots=2 \ \text{(nearest year)}$

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

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

$\Rightarrow Q=950$

♦ Mean mark (g) 40%.

Calculus, SPEC2 2023 VCAA 3

The curve given by $y^2=x-1$, where $2 \leq x \leq 5$, is rotated about the $x$-axis to form a solid of revolution.

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

--- 2 WORK AREA LINES (style=lined) ---
ii. Find the volume of the solid of revolution. (1 mark)

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

--- 4 WORK AREA LINES (style=lined) ---
ii. Hence or otherwise, find the curved surface area of the solid correct to three decimal places. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

The total surface area of the solid consists of the curved surface area plus the areas of the two circular discs at each end.

The 'efficiency ratio' of a body is defined as its total surface area divided by the enclosed volume.

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

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

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

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

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

$V=\dfrac{15 \pi}{2}\ \text{u}^3$

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

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

b.ii. $\text{Evaluate integral in b.i.}$

$\text {S.A.}=30.847\ \text{u}^2$

c. $y=\sqrt{x-1}$

$\text{At}\ \ x=2\ \ \Rightarrow y=1$

$\text{At}\ \ x=5\ \ \Rightarrow y=2$

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

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

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

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

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

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

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

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

Show Worked Solution

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

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

$V=\dfrac{15 \pi}{2}\ \text{u}^3$

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

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

b.ii. $\text{Evaluate integral in b.i.}$

$\text {S.A.}=30.847\ \text{u}^2$

c. $y=\sqrt{x-1}$

$\text{At}\ \ x=2\ \ \Rightarrow y=1$

$\text{At}\ \ x=5\ \ \Rightarrow y=2$

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

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

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

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

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

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

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

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

Complex Numbers, SPEC2 2023 VCAA 2

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

Verify that $w$ is a root of $z^7-1=0$. (1 marks)

--- 2 WORK AREA LINES (style=lined) ---

List the other roots of $z^7-1=0$ in polar form. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

On the Argand diagram below, plot and label the points that represent all the roots of $z^7-1=0$. (2 marks)

--- 4 WORK AREA LINES (style=blank) ---

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)

--- 2 WORK AREA LINES (style=lined) ---

Verify that the equation $z^7-1=0$ can be expressed in the form $(z-1)\left(z^6+z^5+z^4+z^3+z^2+z+1\right)=0$.
--- 4 WORK AREA LINES (style=lined) ---
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 WORK AREA LINES (style=lined) ---
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 $\cos \left(\dfrac{2 \pi}{7}\right)+\cos \left(\dfrac{4 \pi}{7}\right)+\cos \left(\dfrac{6 \pi}{7}\right)=-\dfrac{1}{2}$.
  --- 5 WORK AREA LINES (style=lined) ---

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{)} $

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. $\text{Roots of}\ \ z^7-1=0:$

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.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 $

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

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

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.

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

--- 2 WORK AREA LINES (style=lined) ---

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

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

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

--- 2 WORK AREA LINES (style=lined) ---

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

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

--- 4 WORK AREA LINES (style=lined) ---
The manufacturer would like to improve processes to ensure that more than 99% of all tennis balls produced are classed as grade A.

Assuming that the mean diameter of the tennis balls remains the same, find the required standard deviation of the diameter, in centimetres, correct to two decimal places. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
An inspector takes a random sample of 32 tennis balls from the manufacturer and determines a confidence interval for the population proportion of grade A balls produced.
The confidence interval is (0.7382, 0.9493), correct to four decimal places.

Find the level of confidence that the population proportion of grade A balls is within the interval, as a percentage correct to the nearest integer. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

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

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

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

--- 2 WORK AREA LINES (style=lined) ---
Find the exact mean serving speed for grade A balls, in metres per second. (1 mark)
--- 3 WORK AREA LINES (style=lined) ---

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

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

If the mean serving speed for a grade B ball is $2\pi^2+8$ metres per second, find the values of $a$ and $b$. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---

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, SPEC2 2023 VCAA 1

Viewed from above, a scenic walking track from point $O$ to point $D$ is shown below. Its shape is given by

$f(x)= \begin{cases}-x(x+a)^2, & 0 \leq x \leq 1 \\ e^{x-1}-x+b, & 1<x \leq 2 .\end{cases}$

The minimum turning point of section $O A B C$ occurs at point $A$. Point $B$ is a point of inflection and the curves meet at point $C(1,0)$. Distances are measured in kilometres.

Show that $a=-1$ and $b=0$. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Verify that the two curves meet smoothly at point $C$. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
i. Find the coordinates of point $A$. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
ii. Find the coordinates of point $B$. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---

The return track from point $D$ to point $O$ follows an elliptical path given by

$x=2 \cos (t)+2, y=(e-2) \sin (t)$, where $t \in\left[\dfrac{\pi}{2}, \pi\right]$.

Find the Cartesian equation of the elliptical path. (2 marks)

--- 3 WORK AREA LINES (style=lined) ---
Sketch the elliptical path from $D$ to $O$ on the diagram above. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
i. Write down a definite integral in terms of $t$ that gives the length of the elliptical path from $D$ to $O$. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
ii. Find the length of the elliptical path from $D$ to $O$.
Give your answer in kilometres correct to three decimal places. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $\text{At}\ C(1,0): $

$\underset{x \to 1^{-}}{\lim} f_1(x) = \underset{x \to 1^{-}}{\lim}-x(x+a)^2 = -1(1+a)^2 = 0 $

$a=-1$

$\underset{x \to 1^{+}}{\lim} f_2(x)\ \underset{x \to 1^{+}}{\lim} (e^{x-1}-x+b) = e^0-1+b=b = 0 $

b. $\text{Find derivative functions (by calc)}:$

$\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_1^{′}(x)=0 $

$\dfrac{d}{dx} (e^{x-1}-x+b) = e^{x-1}-1\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_2^{′}(x)=0 $

$\Rightarrow\ \text{Since gradients are equal, curves meet smoothly.}$

c.i. $A\Big{(} \frac{1}{3}, -\frac{4}{27} \Big{)} $

c.ii. $B\Big{(} \frac{2}{3}, -\frac{2}{27} \Big{)} $

d. $\Big{(}\dfrac{x-2}{2}\Big{)}^2 + \Big{(}\dfrac{y}{e-2}\Big{)}^2=1$

f.i. $\displaystyle{\int_{\frac{\pi}{2}}^{\pi}} \sqrt{4\,\sin^2(t)+(e-2)^2\cos^2(t)}\,dt$

f.ii. $\text{Length}\ = 2.255$

Show Worked Solution

a. $\text{At}\ C(1,0): $

$\underset{x \to 1^{-}}{\lim} f_1(x) = \underset{x \to 1^{-}}{\lim}-x(x+a)^2 = -1(1+a)^2 = 0 $

$a=-1$

$\underset{x \to 1^{+}}{\lim} f_2(x)\ \underset{x \to 1^{+}}{\lim} (e^{x-1}-x+b) = e^0-1+b=b = 0 $

b. $\text{Find derivative functions (by calc)}:$

$\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_1^{′}(x)=0 $

$\dfrac{d}{dx} (e^{x-1}-x+b) = e^{x-1}-1\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_2^{′}(x)=0 $

$\Rightarrow\ \text{Since gradients are equal, curves meet smoothly.}$

c.i. $\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \Rightarrow \ \text{SP when}\ x=\dfrac{1}{3} $

$A\Big{(} \frac{1}{3}, -\frac{4}{27} \Big{)} $

c.ii. $\text{POI at}\ B: \ f_1^{″}(x)=-6x+4=0\ \ \Rightarrow x=\dfrac{2}{3} $

$B\Big{(} \frac{2}{3}, -\frac{2}{27} \Big{)} $

d. $\dfrac{x-2}{2}=\cos(t), \ \ \dfrac{y}{e-2}=\sin(t) $

$\Big{(}\dfrac{x-2}{2}\Big{)}^2 + \Big{(}\dfrac{y}{e-2}\Big{)}^2=1$

f.i. $\dfrac{dx}{dt} = -2\sin(t),\ \ \dfrac{dy}{dt} = (e-2)\cos(t) $

$\displaystyle{\int_{\frac{\pi}{2}}^{\pi}} \sqrt{4\,\sin^2(t)+(e-2)^2\cos^2(t)}\,dt$

f.ii. $\text{Evaluate the integral in part f.i.}$

$\Rightarrow \text{Length}\ = 2.255$

PHYSICS, M6 2023 VCE 7*

Two high-voltage transmission lines span a distance of 260 km between Power Plant A and Town B, as shown in Figure 7. Power Plant A provides 350 MW of power. The potential difference at Power Plant A is 500 kV. The current in the transmission lines has a value of 700 A and the power loss in the transmission lines is 20 MW.

Show, using calculations, that the total resistance of the two transmission lines is 41 $\Omega$. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---

Town B needs a minimum of 480 kV.
Determine whether 480 kV will be available to Town B. Show your working. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---

Explain what would happen if the electricity between Power Plant A and Town B were to be transmitted at 50 kV instead of 500 kV. Assume that the resistance of the transmission lines is still 41 $\Omega$. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $41\ \Omega$

b. The power generated in the power plant is 500 kV.

$V_{\text{drop}}=IR=700 \times 41 =28.7\ \text{kV} $

$V_{\text{remaining}} = 500-28.7 = 471.3\ \text{kV} $

$\therefore$ 480V will not be available to Town B.

c. If the voltage was decreased by a factor of 10 to 50 kV:

→ The current through the power lines would increase by a factor of 10.

→ Power remains constant $(P=VI)$

→ As $P_{\text{loss}}=I^2r$, this would dramatically increase the power loss and significantly less power would be transmitted to the town.

Show Worked Solution

a.	$P_{\text{loss}}$	$=I^2R$
	$R$	$=\dfrac{P_{\text{loss}}}{I^2}$
		$=\dfrac{20 \times 10^6}{700^2}$
		$=41\ \Omega$

b. The power generated in the power plant is 500 kV.

$V_{\text{drop}}=IR=700 \times 41 =28.7\ \text{kV} $

$V_{\text{remaining}} = 500-28.7 = 471.3\ \text{kV} $

$\therefore$ 480V will not be available to Town B.

c. If the voltage was decreased by a factor of 10 to 50 kV:

→ Power remains constant $(P=VI)$

→ The current through the power lines would increase by a factor of 10.

→ As $P_{\text{loss}}=I^2r$, this would dramatically increase the power loss and significantly less power would be transmitted to the town.

PHYSICS, M6 2023 VCE 6

Kim and Charlie are attempting to create a DC generator and have arranged the magnets along the axis of rotation of the wire loop, J, K, L and M, as shown in Figure 6. They are having some trouble getting it to work. They rotate the loop in the direction of the arrow, as shown in Figure 6.

Using physics concepts, explain why this orientation of the magnets will not generate an EMF. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Kim and Charlie decide to move the magnets so that an EMF is generated. On Figure 6, draw the positions of the magnets to ensure that an EMF is generated. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a. → The generator has the magnetic field parallel to the plane of the loop.

→ Therefore, there is no flux cutting through the loop of wire.

→ Further, as the loop rotates there is no change in flux and no current or EMF produced.

b. → The magnets should be rotated by 90°.

Show Worked Solution

a. → The generator has the magnetic field parallel to the plane of the loop.

→ Therefore, there is no flux cutting through the loop of wire.

→ Further, as the loop rotates there is no change in flux and no current or EMF produced.

♦ Mean mark (a) 46%.

b. → The magnets should be rotated by 90°.

PHYSICS, M7 2023 VCE 14 MC

Polarisation of visible light provides evidence that electromagnetic radiation can be explained using a

standing wave model for light.
transverse wave model for light.
mechanical wave model for light.
longitudinal wave model for light.

Show Answers Only

$B$

Show Worked Solution

→ For polarisation to occur, the plane of vibration of the wave must be perpendicular to direction of motion.

→ This is a property of the transverse wave model of light.

$\Rightarrow B$

Graphs, MET2 2023 VCAA 1 MC

The amplitude, $A$, and the period, $P$, of the function $f(x)=-\dfrac{1}{2}\sin(3x+2\pi)$ are

$A=-\dfrac{1}{2},\ P=\dfrac{\pi}{3}$
$A=-\dfrac{1}{2},\ P=\dfrac{2\pi}{3}$
$A=-\dfrac{1}{2},\ P=\dfrac{3\pi}{2}$
$A=\dfrac{1}{2},\ P=\dfrac{\pi}{3}$
$A=\dfrac{1}{2},\ P=\dfrac{2\pi}{3}$

Show Answers Only

$E$

Show Worked Solution

$\text{Period:}\ \ P= \dfrac{2\pi}{n}= \dfrac{2\pi}{3}$

$\text{Amplitude}:\ \ A=\bigg|-\dfrac{1}{2}\bigg|=\dfrac{1}{2}$

$\Rightarrow E$

Calculus, MET1 2023 VCAA 9

The shapes of two walking tracks are shown below.

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

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

The unit of length is kilometres.

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

--- 12 WORK AREA LINES (style=lined) ---

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.

Calculus, MET1 2023 VCAA 7

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

State the range of $f$. (1 mark)
--- 2 WORK AREA LINES (style=lined) ---
Sketch the graph of the inverse function $y=f^{-1}(x)$ on the axes above. Label any endpoints and axial intercepts with their coordinates. (2 marks)
--- 0 WORK AREA LINES (style=lined) ---
Determine the equation of the domain for the inverse function $f^{-1}$. (2 marks)
--- 3 WORK AREA LINES (style=lined) ---
Calculate the area of the regions enclosed by the curves of $f,\ f^{-1}$ and $y=-x$. (2 marks)
--- 8 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $[-1, \infty)$

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

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

d. $A=\dfrac{1}{3}$

Show Worked Solution

a. $[-1, \infty)$

c. $\text{When }f(x)\ \text{is written in turning point form}$

$y=(x-1)^2-1$

$\text{Inverse function: swap}\ x \leftrightarrow y$

$x$	$=(y-1)^2-1$
$x+1$	$=(y-1)^2$
$-\sqrt{x+1}$	$=y-1$
$f^{-1}(x)$	$=1-\sqrt{x+1}$

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

♦ Mean mark (c) 48%.
MARKER’S COMMENT: Common error → writing the function as $f^{-1}(x)=1+\sqrt{x+1}$.

d. $\text{One strategy of many possibilities:}$

	$A$	$=2\displaystyle \int_{0}^{1} \big(-x-(x^2-2x)\big)\,dx$
		$=2\displaystyle \int_{0}^{1} \big(x-x^2\big)\,dx$
		$=2\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1$
		$=2\bigg(\dfrac{1}{2}-\dfrac{1}{3}-(0)\bigg)$
		$=\dfrac{1}{3}$

♦♦♦ Mean mark (d) 24%.
MARKER’S COMMENT: Using the symmetry properties of the graph and its inverse helped answer this question efficiently.

Graphs, MET1 2023 VCAA 3

Sketch the graph of $f(x)=2-\dfrac{3}{x-1}$ on the axes below, labelling all asymptotes with their equation and axial intercepts with their coordinates. (3 marks)

--- 0 WORK AREA LINES (style=lined) ---
Find the values of $x$ for which $f(x)\leq1$. (1 mark)
--- 2 WORK AREA LINES (style=lined) ---

Show Answers Only

b. $1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]$

Show Worked Solution

a. $\text{Vertical asymptote when}\ \ x=1$

$y\text{-int:}\ y=2-(-3)\ \ \Rightarrow\ \ y=5$

$x\text{-int:}\ 2-\dfrac{3}{x-1}=0\ \ \Rightarrow\ \ x=\dfrac{5}{2} $

$\text{As}\ \ x \rightarrow \infty, \ \ y \rightarrow 2^{-}; \ \ x \rightarrow -\infty, \ \ y \rightarrow 2^{+} $

b. $\text{From the graph:}$

$f(x)=1\ \text{when }x=4$

$x>1\ \text{to the right of the vertical asymptote}$

$\therefore\ f(x)\leq1\ \text{when}\ \ 1<x\leq4\ \ \ \text{or}\ \ \big(1,4\big]$

♦♦ Mean mark (b) 38%.
MARKER’S COMMENT: Many students did not use their graph from part (a).

Calculus, MET1 2023 VCAA 1b

Let $f(x)=\sin(x)e^{2x}$.

Find $f^{'}\Big(\dfrac{\pi}{4}\Big)$. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

$\dfrac{3\sqrt{2}}{2}e^{\frac{\pi}{2}}\ \text{or}\ \dfrac{3e^{\frac{\pi}{2}}}{\sqrt{2}}$

Show Worked Solution

$\text{Using the product rule}$

$f'(x)$	$=e^{2x}\cos(x)+2e^{2x}\sin(x)$
	$=e^{2x}\Big(\cos(x)+2\ \sin(x)\Big)$
$\therefore\ f’\Big(\dfrac{\pi}{4}\Big)$	$=e^{2(\frac{\pi}{4})}\Bigg(\cos(\dfrac{\pi}{4})+2\ \sin(\dfrac{\pi}{4})\Bigg)$
	$=e^{\frac{\pi}{2}}\Bigg(\dfrac{1}{\sqrt{2}}+\sqrt{2}\Bigg)$
	$=e^{\frac{\pi}{2}}\Bigg(\dfrac{1+\sqrt2 \times \sqrt2}{\sqrt2} \Bigg) $
	$=\dfrac{3\sqrt{2}}{2}e^{\frac{\pi}{2}}\ \ \text{or}\ \ \dfrac{3e^{\frac{\pi}{2}}}{\sqrt{2}}$

Complex Numbers, SPEC2 2023 VCAA 4 MC

If $z=-(2 a+1)+2 a i$, where $a$ is a non-zero real constant, then $\dfrac{4 a}{1+\bar{z}}$ is equal to

$\sqrt{2} \text{cis}\left(\dfrac{\pi}{4}\right)$
$\sqrt{2} \text{cis}\left(\dfrac{3 \pi}{4}\right)$
$\text{cis}\left(\dfrac{\pi}{4}\right)$
$\sqrt{2} \text{cis}\left(-\dfrac{3 \pi}{4}\right)$
$\text{cis}\left(-\dfrac{\pi}{4}\right)$

Show Answers Only

$B$

Show Worked Solution

$\text{Define}\ z\ \text{(by CAS):} $

$\dfrac{4 a}{1+\bar{z}} = -1+i $

$\Rightarrow B$

Functions, SPEC1 2023 VCAA 1

Consider the function $f$ with rule $f(x)=\dfrac{x^2+x-6}{x-1}$.

Show that the rule for the function $f$ can be written as $f(x)=x+2-\dfrac{4}{x-1}$. (1 mark)

--- 6 WORK AREA LINES (style=lined) ---

Sketch the graph of $f$ on the axes below, labelling any asymptotes with their equations. (3 marks)

--- 0 WORK AREA LINES (style=blank) ---

Show Answers Only

a.	$f(x)$	$=\dfrac{x^2+x-6}{x-1}$
		$=\dfrac{(x-1)(x+2)-4}{x-1}$
		$=x+2-\dfrac{4}{x-1} $

b. $\text{Asymptotes at:}\ x=1, \ y=x+2 $

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -1 & 0 & 1 & 2 \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 0 & 3 & 6 & \infty & 0 \\
\hline
\end{array}

Show Worked Solution

a.	$f(x)$	$=\dfrac{x^2+x-6}{x-1}$
		$=\dfrac{(x-1)(x+2)-4}{x-1}$
		$=x+2-\dfrac{4}{x-1} $

b. $\text{Asymptotes at:}\ x=1, \ y=x+2 $

\begin{array} {|l|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -3 & -1 & 0 & 1 & 2 \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & 0 & 3 & 6 & \infty & 0 \\
\hline
\end{array}

Statistics, SPEC1 2023 VCAA 6

Josie travels from home to work in the city. She drives a car to a train station, waits, and then rides on a train to the city. The time, $X_c $ minutes, taken to drive to the station is normally distributed with a mean of 20 minutes $ (\mu_c=20) $ and standard deviation of 6 minutes $(\sigma_c=6) $. The waiting time, $ X_w $ minutes, for a train is normally distributed with a mean of 8 minutes $ (\mu_w=8) $ and standard deviation of $ \sqrt{3} $ minutes $ (\sigma_w=\sqrt{3}) $. The time, $ X_t $ minutes, taken to ride on a train to the city is also normally distributed with a mean of 12 minutes $ (\mu_t=12) $ and standard deviation of 5 minutes $ (\sigma_t=5) $. The three times are independent of each other.

Find the mean and standard deviation of the total time, in minutes, it takes for Josie to travel from home to the city. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Josie's waiting time for a train on each work day is independent of her waiting time for a train on any other work day. The probability that, for 12 randomly chosen work days, Josie's average waiting time is between 7 minutes 45 seconds and 8 minutes 30 seconds is equivalent to $ \text{Pr}(a<Z<b)$, where $Z \sim \text{N}(0,1)$ and $a$ and $b$ are real numbers.
Find the values of $a$ and $b$. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $E[X_c+X_w+X_t]=40$

$\sigma[X_c+X_w+X_t]=8$

b. $a=-0.5, \ b=1$

Show Worked Solution

a. $X_c ∼ N(20, 6^2), X_w ∼ N(8, (\sqrt{3})^2), X_t ∼ N(12, 5^2) $

$E[X_c+X_w+X_t]$	$=E[X_c]+E[X_w]+E[X_t]$
	$=20+8+12$
	$=40$

$\text{Var}[X_c+X_w+X_t]$	$=\text{Var}[X_c]+\text{Var}[X_w]+\text{Var}[X_t]$
	$=36+3+25$
	$=64$
$\sigma[X_c+X_w+X_t]$	$=8$

b. $E(\bar X)=\mu_w=8,\ \ \sigma(\bar X)=\dfrac{\sigma_w}{\sqrt{12}}=\dfrac{\sqrt3}{\sqrt{12}}=\dfrac{1}{2} $

$\text{Pr}(7.75<\bar X<8.5)$	$=\text{Pr} \Bigg{(} \dfrac{7.75-8}{\frac{1}{2}}<Z<\dfrac{8.5-8}{\frac{1}{2}}\Bigg{)} $
	$=\text{Pr} \Bigg{(} \dfrac{-0.25}{\frac{1}{2}}<Z<\dfrac{0.5}{\frac{1}{2}}\Bigg{)} $
	$=\text{Pr}(-0.5<Z<1)$

$\therefore a=-0.5, \ b=1$

Calculus, SPEC1 2023 VCAA 5

Evaluate $
\begin{aligned}
\int_1^2x^2\, \log_e(x)\ dx. \\
\end{aligned}
$ (3 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

$\dfrac{8}{3}\log_e(2)-\dfrac{7}{3} $

Show Worked Solution

	$u$	$= \log_e x,$	$\ \ \ \ u^{′}$	$= \dfrac{1}{x}$
	$v^{′}$	$= x^2,$	$v$	$= \dfrac{1}{3} x^3$

\[\int_1^2 x^2\,\log_e (x) \,dx\]	\[= uv-\int u^{′} v \ dx\]
	\[= \Big{[}\dfrac{x^3}{3}\,\log_e x\Big{]}_1 ^2-\dfrac{1}{3} \int_1 ^2 \dfrac{1}{x} \times x^3\ dx\]
	\[= \Big{[} \dfrac{8}{3} \log_e 2-0\Big{]}-\dfrac{1}{9} \Big{[}x^3\Big{]}_1^2\]
	$=\dfrac{8}{3} \log_e 2-\dfrac{1}{6}(8-1) $
	$=\dfrac{8}{3} \log_e 2-\dfrac{7}{9} $

Networks, GEN2 2023 VCAA 12

A country has five states, $A, B, C, D$ and $E$.

A graph can be drawn with vertices to represent each of the states.

Edges represent a border shared between two states.

What is the sum of the degrees of the vertices of the graph above? (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
Euler's formula, $v+f=e+2$, holds for this graph.
1. Complete the formula by writing the appropriate numbers in the boxes provided below. (1 mark)
  
  --- 0 WORK AREA LINES (style=lined) ---
2. Complete the sentence by writing the appropriate word in the space provided below. (1 mark)
  
  --- 0 WORK AREA LINES (style=lined) ---
3. Euler’s formula holds for this graph because the graph is connected and ______________.
The diagram below shows the position of state $A$ on a map of this country.
The four other states are indicated on the diagram as 1, 2, 3 and 4.

Use the information in the graph above to complete the table below. Match the state $(B, C, D$ and $E)$ with the corresponding state number $(1,2,3$ and 4$)$ given in the map above. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \textbf{State} \quad \quad \rule[-1ex]{0pt}{0pt} & \textbf{State Number} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} &\\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}

Show Answers Only

a. $\text{Sum of degrees}\ = 2+3+4+3+2 = 14$

b.i. $\text{Vertices = 5, Faces = 4, Edges = 7}$

$\Rightarrow 5 + 4 = 7 + 2$

b.ii. $\text{Planar}$

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \textbf{State} \quad \quad \rule[-1ex]{0pt}{0pt} & \textbf{State Number} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 3\\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 2 \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & 4 \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & 1 \\
\hline
\end{array}

Show Worked Solution

a. $\text{Sum of degrees}\ = 2+3+4+3+2 = 14$

b.i. $\text{Vertices = 5, Faces = 4, Edges = 7}$

$\Rightarrow 5 + 4 = 7 + 2$

b.ii. $\text{Planar}$

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \quad \quad \textbf{State} \quad \quad \rule[-1ex]{0pt}{0pt} & \textbf{State Number} \\
\hline
\rule{0pt}{2.5ex} B \rule[-1ex]{0pt}{0pt} & 3\\
\hline
\rule{0pt}{2.5ex} C \rule[-1ex]{0pt}{0pt} & 2 \\
\hline
\rule{0pt}{2.5ex} D \rule[-1ex]{0pt}{0pt} & 4 \\
\hline
\rule{0pt}{2.5ex} E \rule[-1ex]{0pt}{0pt} & 1 \\
\hline
\end{array}

Recursion and Finance, GEN2 2023 VCAA 5

Arthur borrowed $30 000 to buy a new motorcycle.

Interest on this loan is charged at the rate of 6.4% per annum, compounding quarterly.

Arthur will repay the loan in full with quarterly repayments over six years.

How many repayments, in total, will Arthur make? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---

The balance of the loan, in dollars, after $n$ quarters, $A_n$, can be modelled by the recurrence relation

$A_0=30\ 000, \quad A_{n+1}=1.016 A_n-1515.18$

Showing recursive calculations, determine the balance of the loan after two quarters.
Round your answer to the nearest cent. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
The final repayment required will differ slightly from all the earlier repayments of $1515.18
Determine the value of the final repayment.
Round your answer to the nearest cent. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $\text{Repayments}\ = 6 \times 4 = 24$

b. $A_1=1.016 \times 30\ 000-1515.18=$28\ 964.82 $

$A_2=1.016 \times 28\ 964.82-1515.18=$27\ 913.08 $

c. $ \text{Final repayment}\ = 1515.18-0.14=\$1515.04$

Show Worked Solution

a. $\text{Repayments}\ = 6 \times 4 = 24$

b. $A_1=1.016 \times 30\ 000-1515.18=$28\ 964.82 $

$A_2=1.016 \times 28\ 964.82-1515.18=$27\ 913.08 $

Mean mark (b) 51%.
♦♦ Mean mark (c) 25%

c. $\text{Solve for}\ N\ \text{using TMV calculator:}$

$N$	$=24$
$I\%$	$=6.4$
$PV$	$= -30\ 000$
$PMT$	$= 1515.18$
$FV$	$=?$
$\text{P/Y}$	$=\ \text{C/Y}\ = 4$

$FV = -0.14$

$\therefore \text{Final repayment}\ = 1515.18-0.14=\$1515.04$

Data Analysis, GEN2 2023 VCAA 3

The scatterplot below plots the average monthly ice cream consumption, in litres/person, against average monthly temperature, in °C. The data for the graph was recorded in the Northern Hemisphere.

When a least squares line is fitted to the scatterplot, the equation is found to be:

consumption = 0.1404 + 0.0024 × temperature

The coefficient of determination is 0.7212

Draw the least squares line on the scatterplot graph above. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
Determine the value of the correlation coefficient $r$.
Round your answer to three decimal places. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Describe the association between average monthly ice cream consumption and average monthly temperature in terms of strength, direction and form. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{strength} \rule[-1ex]{0pt}{0pt} & \quad \quad \quad \quad \quad \quad \quad \quad \\
\hline
\rule{0pt}{2.5ex} \textbf{direction} \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} \textbf{form} \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}
Referring to the equation of the least squares line, interpret the value of the intercept in terms of the variables consumption and temperature. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Use the equation of the least squares line to predict the average monthly ice cream consumption, in litres per person, when the monthly average temperature is –6°C. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Write down whether this prediction is an interpolation or an extrapolation. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---

Show Answers Only

b. $r = \sqrt{0.7212} = 0.849\ \text{(3 d.p.)}$

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \textbf{strength} \rule[-1ex]{0pt}{0pt} & \text{strong} \\
\hline
\rule{0pt}{2.5ex} \textbf{direction} \rule[-1ex]{0pt}{0pt} & \text{positive} \\
\hline
\rule{0pt}{2.5ex} \textbf{form} \rule[-1ex]{0pt}{0pt} & \text{linear} \\
\hline
\end{array}

d. $\text{At 0°C, the predicted average consumption is:}$

$\textit{consumption} = 0.1404 + 0.002 \times 0 = 0.1404\ \text{L/person}$

e. $\text{Find consumption} (c)\ \text{when temperature} (t) = -6:$

$c=0.1404 + 0.002 \times -6 = 0.1284\ \text{L/person} $

f. $\text{Extrapolation (even though the axes extend to –6°C, the data set} $

$\text{range finishes with a lower limit around –4.5°C.)}$

Show Worked Solution

b. $r = \sqrt{0.7212} = 0.849\ \text{(3 d.p.)}$

d. $\text{At 0°C, the predicted average consumption is:}$

$\textit{consumption} = 0.1404 + 0.002 \times 0 = 0.1404\ \text{L/person}$

e. $\text{Find consumption} (c)\ \text{when temperature} (t) = -6:$

$c=0.1404 + 0.002 \times -6 = 0.1284\ \text{L/person} $

f. $\text{Extrapolation (even though the axes extend to –6°C, the data set} $

$\text{range finishes with a lower limit around –4.5°C.)}$

« Previous Page
1
…
3
4
5
6
7
…
42
Next Page »

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

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

b.	\(x_1\)	\(=x_0-\dfrac{f(x_0)}{f^{\prime}(x_0)},\quad\ \text{where }x_0=-1\)
		\(=-1-\dfrac{-\dfrac{1}{3}+2.(-1)+4}{(-1)^2+2}\)
		\(=-\dfrac{14}{9}\)

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

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

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

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

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

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