Band 4

Calculus, 2ADV C3 2025 MET2 19*

Let $A$ be a point on the line $y=x+c$ and $B$ be a point on the curve $y=\log _e(x-1)$.

The tangent to the curve at point $B$ is parallel to the line $y=x+c$.

Show that the distance $(d)$ between the points $AB$ can be expressed as
$d=\sqrt{2x^2+(2c-4)x+4+c^2} $ (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Determine the $x$-coordinate of point $A$, in terms of $c$, when the distance $AB$ is a minimum. (2 marks)

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

Show Answers Only

a. $\text{See Worked Solutions}$

b. $d_{\text{min}}=\dfrac{2-c}{2}$

Show Worked Solution

a. $\text{Gradient of}\ \ y=x+c \ \ \Rightarrow\ \ m=1$

$g(x)=\log _e(x-1), \ g^{\prime}(x)=\dfrac{1}{x-1}$

$\text{Solve} \ \ g^{\prime}(x)=1:$

$\dfrac{1}{x-1}=1 \ \Rightarrow \ x=2$

$\text{Consider the diagram for the case when}\ \ c=0:$

$\text{Find distance \((d)$ between $B(2,0)$ and $A(x, x+c)$:}\)

$d$	$=\sqrt{(x-2)^2+(x+c)^2}$
	$=\sqrt{x^2-4x+4+x^2+2cx+c^2}$
	$=\sqrt{2x^2+(2c-4)x+4+c^2}$

b. $d^2=2x^2+(2c-4)x+4+c^2$

$\dfrac{d(d^2)}{dx}=4x+2c-4$

$\dfrac{d^2(d^2)}{dx^2}=4>0$

$\text{MIN when}\ \dfrac{d(d^2)}{dx}=0:$

$4x+2c-4=0\ \ \Rightarrow\ \ x=\dfrac{4-2c}{4}=\dfrac{2-c}{2}$

$\therefore d_{\text{min}}\ \text{occurs when}\ \ x=\dfrac{2-c}{2}$

Statistics, MET2 2025 VCAA 14 MC

Let $f$ be the probability density function for a continuous random variable $X$, where

\begin{align*}
f(x)=\left\{\begin{array}{cl}
k\, \sin (x) & 0 \leq x<\dfrac{\pi}{4} \\
k\, \cos (x) & \dfrac{\pi}{4} \leq x \leq \dfrac{\pi}{2} \\
0 & \text {otherwise }
\end{array}\right.
\end{align*}

and $k$ is a positive real number.

The value of $k$ is

$\dfrac{1}{\sqrt{2}}$
$\dfrac{1}{2-\sqrt{2}}$
$\sqrt{2}+2$
$2-\sqrt{2}$

Show Answers Only

$B$

Show Worked Solution

$\text{Solve for} \ k \ \text{by (CAS):}$

$\displaystyle \int_{-\infty}^{\infty} f(x)\, d x=1$

$k=\dfrac{\sqrt{2}+2}{2} \times \dfrac{\sqrt{2}-2}{\sqrt{2}-2}=\dfrac{1}{2-\sqrt{2}}$

$\Rightarrow B$

Statistics, MET2 2025 VCAA 12 MC

For a normal random variable $X$, it is known that $\operatorname{Pr}(X>200)=0.325$ and $\operatorname{Pr}(180<X<200)=0.589$

The mean and standard deviation of $X$ are closest to

190 and 10
190 and 11
195 and 10
195 and 11

Show Answers Only

$D$

Show Worked Solution

$\operatorname{Pr}(X<200)=0.325 \ \Rightarrow \ \dfrac{200-\mu}{\sigma}=0.453\ \ldots\ (1)$

$\operatorname{Pr}(180<X<200)=0.589$

$\operatorname{Pr}(X>180)=0.589+0.325=0.914$

$\dfrac{180-\mu}{\sigma}=-1.365\ \ldots\ (2)$

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

$\mu=195, \sigma=11$

$\Rightarrow D$

Mean mark 58%.

Calculus, MET2 2025 VCAA 11 MC

The chart below shows the daily price of a stock market share over a 30-day period.

Over which of the following time intervals did the daily price undergo the greatest average rate of change?

day 3 to day 10
day 3 to day 17
day 14 to day 21
day 14 to day 28

Show Answers Only

$D$

Show Worked Solution

$\text{Day 14 to Day 28 is the steepest.}$

$\text{Average ROC} \approx \dfrac{39.4-35.0}{28-14} \approx 0.3143$

$\Rightarrow D$

Functions, MET2 2025 VCAA 10 MC

Consider $f: R \rightarrow R, f(x)=2 x^2+x-1$ and $g: R \rightarrow R, g(x)=\sin (x)$.

The inequality $(f \circ g)(x)>0$ is satisfied when

$\quad \sin (x) \leq-1$
$-1<\sin (x)<0$
$\dfrac{1}{2}<\sin (x) \leq 1$
$0<\sin (x)<\frac{1}{2}$

Show Answers Only

$C$

Show Worked Solution

$f(g(x))=2 \sin ^2(x)+\sin (x)-1$

$\text{Let} \ \ a=\sin (x)$

$f(g(x))=2 a^2+a-1$

$\text {Solve simultaneously for \(a$ (by CAS):}\)

$2 a^2+a-1>0\ …\ (1)$

$-1 \leqslant a \leqslant 1\ …\ (2)$

$\therefore \frac{1}{2}<a \leqslant 1$

$\Rightarrow C$

Mean mark 57%.

ENGINEERING, CS 2025 HSC 11 MC

Which of the following is true of cathodic protection?

It uses high temperatures to preserve metals.
It uses a polymer coating to prevent corrosion.
It uses heat treatment to improve the tensile strength.
It uses an electric current to prevent corrosion of metals.

Show Answers Only

$D$

Show Worked Solution

Cathodic protection works by supplying an electric current that makes the metal structure the cathode, preventing oxidation (corrosion).
A is incorrect — high temperatures accelerate corrosion rather than prevent it.
B describes paint/polymer coating — a separate corrosion prevention method, not cathodic protection.
C describes heat treatment — a process that changes mechanical properties, unrelated to corrosion prevention.

$\Rightarrow D$

ENGINEERING, AE 2025 HSC 10 MC

Which mechanical property of a thermosetting polymer increases when the number of cross-links between the polymer chains is increased?

Ductility
Flexibility
Stiffness
Toughness

Show Answers Only

$C$

Show Worked Solution

Cross-links are covalent bonds between polymer chains — more cross-links restrict chain movement, increasing stiffness.
Ductility and flexibility both require chain movement — cross-linking reduces these properties.
Toughness requires energy absorption through deformation — heavily cross-linked polymers become brittle, not tough.

$\Rightarrow C$

ENGINEERING, PPT 2025 HSC 6 MC

The diagram shows a 100 kg crate which is being pulled horizontally 10 metres along level ground. The tension in the tow rope is 600 N and the rope is inclined at 60 degrees to the ground on a frictionless surface.

What is the work done in joules?

$100 \times 10$
$600 \times 10$
$600 \sin 60^{\circ} \times 10$
$600 \cos 60^{\circ} \times 10$

Show Answers Only

$D$

Show Worked Solution

$W$	$= F \cos\theta \times d$
	$= 600 \cos 60^{\circ} \times 10$
	$= 600 \times 0.5 \times 10$
	$= 3000 \ \text{J}$

$\Rightarrow D$

ENGINEERING, TE 2025 HSC 3 MC

The truth table represents the basic operation of a logic gate.

\begin{array}{|c|c|c|}
\hline
\rule{0pt}{2.5ex}\ \ \ \textit{Input A} \ \ \ \rule[-1ex]{0pt}{0pt}&\ \ \ \textit{Input B} \ \ \ & \ \ \ \textit{Output} \ \ \ \\
\hline
\rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt}& 1 & 1 \\
\hline
\rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt}& 0 & 1 \\
\hline
\rule{0pt}{2.5ex} 0 \rule[-1ex]{0pt}{0pt}& 1 & 1 \\
\hline
\rule{0pt}{2.5ex} 0 \rule[-1ex]{0pt}{0pt}& 0 & 0 \\
\hline
\end{array}

What is the logic gate?

AND
NAND
NOR
OR

Show Answers Only

$D$

Show Worked Solution

An OR gate produces an output of 1 whenever at least one input is 1, and an output of 0 only when both inputs are 0.
The truth table shows Output = 1 for all input combinations except (0, 0), which is the defining behaviour of an OR gate.
An AND gate requires both inputs to be 1 to produce an output of 1 — eliminated by rows where one input is 0 yet the output is 1.
NAND and NOR gates are inverted versions of AND and OR respectively — their outputs would be reversed from those shown.

$\Rightarrow D$

Trigonometry, 2ADV T3 2025 MET2 3 MC

The graph of $y=f(x)$ is shown below.

Which one of the following options best represents the graph of $y=f(-x)+2$ ?

Show Answers Only

$B$

Show Worked Solution

$\text{Transformation:}$

$\text{reflect} \ \ y=f(x) \ \text {in} \ y \text {-axis}$
$\text{translate up 2 units}$

$\Rightarrow B$

Probability, MET2 2025 VCAA 9 MC

One day, at a particular school, $m$ students walked to school and the remaining $n$ students travelled to school using a different form of transport.

Of the $m$ students who walked, 20% took at least 30 minutes to get to school.

Of the $n$ students who used a different form of transport, 40% took at least 30 minutes to get to school.

Given that a randomly selected student took at least 30 minutes to get to school, the probability that they walked to school is given by

$\dfrac{m}{m+2 n}$
$\dfrac{2 n}{m+2 n}$
$\dfrac{m}{5(m+n)}$
$\dfrac{1}{3}$

Show Answers Only

$A$

Show Worked Solution

$\text{Let \(W=$ student walks to school}\)

$\text{Let \(T=$ travel time at least 30 mins}\)

Mean mark 57%.

$P(W)=\dfrac{\text{students who walk}}{\text{total students}} = \dfrac{m}{m+n}$

$P(T)=\dfrac{\text{travel time at least 30 mins}}{\text{total students}} =\dfrac{0.2 m+0.4 n}{m+n}$

$P(W \cap T)=\dfrac{0.2 m}{m+n}$

$P\left(\left.W\right|T\right)=\dfrac{P\left(W \cap T\right)}{P(T)}=\dfrac{\left(\dfrac{0.2 m}{m+n}\right)}{\left(\dfrac{0.2 m+0.4 n}{m+n}\right)}=\dfrac{m}{m+2 n}$

$\Rightarrow A$

Probability, MET2 2025 VCAA 8 MC

A random sample of $n$ Victorian households is taken to estimate the proportion of all Victorian households that have vegetable gardens. The approximate 95% confidence interval calculated using this sample is (0.248, 0.552), correct to three decimal places.

The number of households, $n$, in the sample is

Show Answers Only

$C$

Show Worked Solution

$\text{95% CI} =(0.248,0.552)$

$\hat{p} \approx \dfrac{0.248+0.554}{2}=0.4$

$\text{Solve for} \ n \ \text{(by CAS):}$

$1.96 \times \sqrt{\dfrac{0.4 \times(1-0.4)}{n}} \approx 0.152 \ \Rightarrow \ n \approx 39.905$

$\Rightarrow C$

Calculus, MET2 2025 VCAA 7 MC

Consider the algorithm below.

In order, the values printed by the algorithm are

$ 12$
$ 12,7$
$12,7,2$
$12,7,2,-3$

Show Answers Only

$C$

Show Worked Solution

$n=17, k=5, n>k$

$n=12, k=5, \ \text{print}\ 12, n>k$

$n=7, k=5, \ \text{print} \ 7, n>k$

$n=2, k=5, \ \text{print}\ 2, n<k \text { end while}$

$\text { Printed values:}\ 12,7,2$

$\Rightarrow C$

Statistics, SPEC2 2025 VCAA 6

The volume of water, $V$ mL, consumed by a student during a school day may be assumed to be normally distributed with a mean of 1000 mL and a standard deviation of 80 mL .

1. Write down the mean and standard deviation of the sampling distribution for the average volume of water consumed by randomly selected samples of 25 students.
2. Give your answers in millilitres. (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---
3. What is the probability, correct to four decimal places, that the average volume of water consumed by a random sample of 25 students on a particular school day is more than 970 mL? (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---

The canteen at a particular school stocks two brands of water in bottles, Wasser and Apa.

The manufacturer of Wasser bottled water knows that the volume of water dispensed into bottles may be assumed to be normally distributed with a standard deviation of 5 mL. Engineers at the company take a random sample of 30 bottles and measure the volume of water in each bottle. The sample mean is found to be 750 mL.

Find a 95% confidence interval for the mean volume of water dispensed into each Wasser bottle.
Give your values in millilitres, correct to one decimal place. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
The engineers decide to take 300 random samples, each containing 30 bottles, and calculate the respective 95% confidence intervals. All samples are independent.
In how many of these confidence intervals would the engineers expect the value of the true mean volume dispensed to be included? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
What is the minimum size of the sample required to ensure that the difference between the sample mean and the mean volume dispensed is no more than 1 mL at the 95% confidence level? (1 mark)

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

The volume of water dispensed into Apa water bottles may be assumed to be normally distributed with a mean of 750 mL and a standard deviation of 5 mL. After a service, a random sample of 50 bottles gave a sample mean of 748 mL. The company now claims that the mean volume of water dispensed is less than the stated mean of 750 mL.

A one-tailed statistical test at the 1% level of significance is proposed.

Write down the null and alternative hypotheses that will be used in testing the company's claim. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
1. Determine the $p$ value for this test.
2. Give your answer correct to four decimal places. (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---
3. Is the company's claim correct?
4. Explain your conclusion in terms of the $p$ value. (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---
At the 1% level of significance for a sample size of 50 bottles, find the critical value of the sample mean, below which a sample mean value would support the conclusion that the mean volume of water dispensed is now less than 750 mL.
Give your answer correct to three decimal places. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Assume that, after the service, the true mean volume of water in the Apa bottles was found to be 747.5 mL and that the population standard deviation, $\sigma$, is 5 mL.
At the 1% level of significance, for a sample size of 50 , find the probability that the company will conclude that the service has not reduced the mean volume of water in an Apa bottle.
Give your answer correct to three decimal places. (1 mark)

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

Show Answers Only

a.i. $\mu=1000,\ \ \sigma=16$

a.ii. $P(V>970)=0.9696$

b. $\text{95% CI}=(748.2,751.8)$

c. $95 \% \times 300=285$

d. $n=97$

e. $H_0: \ \mu=750, \ H_1: \ \mu<750$

f.i. $p=0.0023$

f.ii. $\text{Since \(0.0023<0.01$ (significance level), claim is correct.}\)

g. $\overline{X}=748.355 \ \text{mL}$

h. $p=0.113$

Show Worked Solution

a.i.	$\mu$	$=1000$
	$\sigma$	$=\dfrac{80}{\sqrt{16}}=16$

a.ii. $P(V>970)=0.9696$

b. $\overline{X} \sim N\left(750, \dfrac{5}{\sqrt{30}}\right)$

$\text{95% CI}$	$=\left(750-1.96 \times \dfrac{5}{\sqrt{30}}, 750+1.96\times\dfrac{5}{\sqrt{30}}\right)$
	$=(748.2,751.8)$

c. $95 \% \times 300=285$

d. $\text {Solve for} \ n:$

$1$	$\geqslant 1.96 \times \dfrac{5}{\sqrt{n}}$
$n$	$\geqslant 96.04$
$n$	$=97$

e. $H_0: \ \mu=750$

$H_1: \ \mu<750$

f.i.	$p$	$=\operatorname{Pr}\left(z<\dfrac{748-750}{\frac{5}{\sqrt{50}}}\right)$
		$=\operatorname{Pr}\left(z<-\dfrac{2 \sqrt{50}}{5}\right)$
		$=0.0023$

f.ii. $\text{Since \(0.0023<0.01$ (significance level), claim is correct.}\)

g. $\text{Solve for} \ c:\ \ \operatorname{Pr}(\overline{X}<c)=0.01$

$c=748.355 \ \text{mL}$

h. $p=0.113$

Vectors, SPEC2 2025 VCAA 5

Consider three planes defined by the equations $\Pi_1: \ 2 x+9 z=8, \Pi_2: \ 3 x+6 y+5 z=7$ and $\Pi_3: \ x+9 y-3 z=7$.

Find the point of intersection of the three planes. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
1. Find a vector that gives the direction of the line of intersection of the planes $\Pi_2$ and $\Pi_3$. (2 marks)
  
  --- 4 WORK AREA LINES (style=lined) ---
2. Find a set of parametric equations that give the coordinates of the points that lie on this line of intersection. (1 mark)
  
  --- 4 WORK AREA LINES (style=lined) ---
Find the shortest distance from the point $(1,1,2)$ to the plane $\Pi_3$. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Consider a family of planes, $\Psi$, with equation $6 x+27 z=m$, where $m \in N$.
1. Show that the plane $\Pi_1$ is parallel to each member of $\Psi$. (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---
2. Find all values of $m$ for which the shortest distance between plane $\Pi_1$ and the plane of the form $6 x+27 z=m$ is $\dfrac{23}{3 \sqrt{85}}$. (3 marks)
  
  --- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $\text{Intersection at}\ (-5,2,2)$

b.i. $\underset{\sim}{d}=\left(\begin{array}{c}-63 \\ 14 \\ 21\end{array}\right)$

b.ii. $x=-5-63 \lambda$

$y=2 + 14 \lambda$

$z=2+21 \lambda$

c. $\dfrac{3}{\sqrt{91}}$

d.i. $\Psi: \ 6 x+27 z=m$

$\text{Normal to} \ \Psi: \ 6 \underset{\sim}{i}+27 \underset{\sim}{k}$

$\text{Normal to} \ \Pi_2: \ 2 \underset{\sim}{i}+9 \underset{\sim}{k}$

$\displaystyle \binom{6}{27}=3\binom{2}{9} \ \Rightarrow \ \text{Planes are parallel}$

d.ii. $m=1,47$

Show Worked Solution

a. $\Pi_1: \ 2 x+9 z=8$

$\Pi_2: \ 3 x+6 y+5 z=7$

$\Pi_3: \ x+9 y-3 z=7$

$\text {Solve simultaneously (by CAS):}$

$\text{Intersection at}\ (-5,2,2)$

b.i. $\underset{\sim}{d} \ \text{is} \perp \text{to normals of} \ \Pi_2 \ \text{and}\ \Pi_3.$

$\underset{\sim}{d}=n_2 \times n_3=\left|\begin{array}{ccc}i & j & k \\ 3 & 6 & 5 \\ 1 & 9 & -3\end{array}\right|=\left(\begin{array}{c}-63 \\ 14 \\ 21\end{array}\right)$

b.ii. $(-5,2,2)\ \text{lies on both planes (from part a.)}$

$x=-5-63 \lambda$

$y=2 + 14 \lambda$

$z=2+21 \lambda$

♦ Mean mark (b.ii) 51%.

c. $\text{Find shortest distance from (1, 1, 2) to \(\Pi_3$.}\)

$\Pi_3: \ x+9 y-3 z=7 \ \ \Rightarrow \ \ (7,0,0)\ \text{lies on plane.}$

$\text {Spanning vector to} \ (1,1,2) \ \text{is} \ (-6 \underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k})$

$\text{Distance}=\dfrac{1}{\sqrt{91}} \times \abs{\left(\begin{array}{c}-6 \\ 1 \\ 2\end{array}\right)\left(\begin{array}{c}1 \\ 9 \\ -3\end{array}\right)}=\dfrac{3}{\sqrt{91}}$

d.i. $\Psi: \ 6 x+27 z=m$

$\text{Normal to} \ \Psi: \ 6 \underset{\sim}{i}+27 \underset{\sim}{k}$

$\text{Normal to} \ \Pi_2: \ 2 \underset{\sim}{i}+9 \underset{\sim}{k}$

$\displaystyle \binom{6}{27}=3\binom{2}{9} \ \Rightarrow \ \text{Planes are parallel}$

d.ii. $\text{Point on}\ \Pi_1: (4,0,0)$

$\text{Point on} \ \Psi:\left(\dfrac{m}{6}, 0,0\right)$

$\text{Spanning vector}=\left(4-\dfrac{m}{6}\right) \underset{\sim}{i}$

$\text{Distance}=\abs{\left(4-\dfrac{m}{6}\right) \underset{\sim}{i} \cdot \dfrac{(2 \underset{\sim}{i}+9\underset{\sim}{k})}{\sqrt{85}}}=\dfrac{23}{3 \sqrt{85}}$

$\text{Solve}\ \ \abs{8-\dfrac{m}{3}}=\dfrac{23}{3} \ \ \text{for} \ m:$

$m=1,47$

♦ Mean mark (d.ii) 45%.

Vectors, SPEC2 2025 VCAA 4

The path of a moving particle with position vector

$\underset{\sim}{r}(t)=\left(5 \cos (t)-4 \cos \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{i}+\left(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{j}$

is shown below for time $t \geq 0$.

All lengths are in metres and time is measured in seconds.

Write down the coordinates of the particle's starting point. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
On the graph above, draw an arrow from the point $(9,0)$ to indicate the direction of motion of the particle. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
Find the value of $t$ for which the particle will first return to its starting point. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Show that the speed of the particle, in m s$^{-1}$, at time $t$ can be expressed as $\sqrt{125-100 \cos \left(\dfrac{3 t}{2}\right)}$. (3 marks)

--- 9 WORK AREA LINES (style=lined) ---
What is the maximum speed of the particle in m s$^{-1}$? (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
On the graph above, trace the path of the particle for $t \in[0, \pi]$. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
Find the length of the path traced in part f, giving your answer in metres, correct to one decimal place. (2 marks)

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

Show Answers Only

a. $\text{Initial coordinates:}\ (1,0)$

c. $t=4 \pi \ \text{seconds}$

d. $r(t)=\left(5 \cos (t)-4 \cos \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{i}+\left(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{j}$

$\dot{r}(t)=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{i}+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{j}$

$\text {Speed}=\abs{\dot{r}(t)}:$

$\text{Speed}^2$	$=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right)^2+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right)^2$
	$=25 \sin ^2(t)-100 \sin (t) \sin \left(\dfrac{5 t}{2}\right)+100 \sin ^2\left(\dfrac{5 t}{2}\right)$
	$\quad +25 \cos ^2(t)-100 \cos (t) \cos \left(\dfrac{5 t}{2}\right)+100 \cos ^2\left(\dfrac{5 t}{2}\right)$
	$=125-100\left(\sin (t) \sin \left(\dfrac{5 t}{2}\right)-\cos (t) \cos \left(\dfrac{5 t}{2}\right)\right)$
	$=125-100 \cos \left(\dfrac{5 t}{2}-t\right)$
$\text{speed}$	$=\sqrt{125-100 \cos \left(\dfrac{3t}{2}\right)}$

e. $\text{Speed}_{\text {max}}=15 \ \text{ms}^{-1}$

f. $\text{Path traces curve from}\ (1,0)\ \text{to}\ (-5,-4).$

g. $\displaystyle \int_0^\pi \sqrt{125-100 \cos \left(\frac{3 t}{2}\right)} d t=36.6\ \text{(1 d.p.)}$

Show Worked Solution

a. $\text{At} \ \ t=0:$

$r(t)=(5 \cos 0-4 \cos 0)\underset{\sim}{i} + (5 \sin 0-4 \sin 0) {\underset{\sim}{j}}=\underset{\sim}{i}$

$\text{Initial coordinates:}\ (1,0)$

c. $\text{Solve for}\ t \ \text{(by CAS):}$

$5 \cos (t)-4 \cos \left(\dfrac{5t}{2}\right)=1$

$5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)=0$

$t=4 \pi \ \text{seconds}$

Mean mark (c) 53%.

d. $r(t)=\left(5 \cos (t)-4 \cos \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{i}+\left(5 \sin (t)-4 \sin \left(\dfrac{5 t}{2}\right)\right)\underset{\sim}{j}$

$\dot{r}(t)=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{i}+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right) \underset{\sim}{j}$

$\text {Speed}=\abs{\dot{r}(t)}:$

$\text{Speed}^2$	$=\left(-5 \sin (t)+10 \sin \left(\dfrac{5 t}{2}\right)\right)^2+\left(5 \cos (t)-10 \cos \left(\dfrac{5 t}{2}\right)\right)^2$
	$=25 \sin ^2(t)-100 \sin (t) \sin \left(\dfrac{5 t}{2}\right)+100 \sin ^2\left(\dfrac{5 t}{2}\right)$
	$\quad +25 \cos ^2(t)-100 \cos (t) \cos \left(\dfrac{5 t}{2}\right)+100 \cos ^2\left(\dfrac{5 t}{2}\right)$
	$=125-100\left(\sin (t) \sin \left(\dfrac{5 t}{2}\right)-\cos (t) \cos \left(\dfrac{5 t}{2}\right)\right)$
	$=125-100 \cos \left(\dfrac{5 t}{2}-t\right)$
$\text{speed}$	$=\sqrt{125-100 \cos \left(\dfrac{3t}{2}\right)}$

e. $\text{Speed}_{\text {max }} \ \text {occurs when} \ \ \cos \left(\frac{3 t}{2}\right)=-1$

$\text{Speed}_{\text {max}}=\sqrt{125+100}=15 \ \text{ms}^{-1}$

f. $\text{At} \ \ t=\pi, \quad r(\pi)=-5\underset{\sim}{i}-4\underset{\sim}{j}$

$\text{Path traces curve from}\ (1,0)\ \text{to}\ (-5,-4).$

g. $\text{Length of curve (by CAS):}$

$\displaystyle \int_0^\pi \sqrt{125-100 \cos \left(\frac{3 t}{2}\right)} d t=36.6\ \text{(1 d.p.)}$

Graphs, MET2 2025 VCAA 3 MC

The graph of $y=f(x)$ is shown below.

Which one of the following options best represents the graph of $y=f(-x)+2$ ?

Show Answers Only

$B$

Show Worked Solution

$\text{Transformation:}$

$\text{reflect} \ \ y=f(x) \ \text {in} \ \ y \text {-axis}$
$\text{translate up 2 units}$

$\Rightarrow B$

Calculus, SPEC2 2025 VCAA 3

A tank initially contains 5 kg of salt dissolved in 3000 litres of water. Salty water that contains 0.1 kg of salt per litre of water enters the tank at a rate of 20 litres per minute. The solution is kept thoroughly mixed and drains from the tank via a tap at the same rate of 20 litres per minute.

By considering concentration, explain whether the quantity of salt in the tank increases with time. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Let $Q$ denote the quantity of salt, in kilograms, in the tank at time $t$ minutes.
Show that $Q$ satisfies the differential equation $\dfrac{d Q}{d t}=\dfrac{300-Q}{150}$. (1 mark)

--- 4 WORK AREA LINES (style=lined) ---
Using Euler's method with a step size of 15 minutes, find $Q(30)$, the approximate quantity of salt in the tank after 30 minutes.
Give your answer in kilograms, correct to two decimal places. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Use calculus to solve the differential equation $\dfrac{d Q}{d t}=\dfrac{300-Q}{150}$, expressing $Q$ in terms of $t$. (3 marks)

--- 9 WORK AREA LINES (style=lined) ---
What value does the quantity of salt in the tank approach as time approaches infinity?
Give your answer in kilograms. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Find the time taken for the quantity of salt in the tank to reach 100 kg. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
When the quantity of salt in the tank reaches 100 kg , the tap draining the tank is turned off. Assume that the tank does not overflow and there is no change to the inflow rate.
After the tap is turned off, how many minutes does it take for the concentration of salt in the tank to reach $\dfrac{1}{20} \ \text{kg L}^{-1}$? (1 mark)

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

Show Answers Only

a. $\text{Initial concentration}=\dfrac{5}{3000}=0.001\dot{6} \ \text{kg/L}$

$\text{Concentration entering tank}=0.1 \ \text{kg/L}$

$\text{Since} \ \ 0.1>0.001\dot{6}, \ \text{salt concentration increases with time.}$

b. $Q=\text{salt in tank at time}\ t$

$\dfrac{d Q}{d t}=\dfrac{d Q}{d t_{\text {in }}}-\dfrac{d Q}{d t_{\text {out }}}$

$\dfrac{d Q}{d t}=0.1 \times 20-\dfrac{Q}{3000} \times 20=2-\dfrac{Q}{150}=\dfrac{300-Q}{150}$

c. $61.05 \ \text{kg}$

d. $Q=300-295 e^{-\tfrac{t}{150}}$

e. $\text{As} \ \ t \rightarrow \infty, Q \rightarrow 300$

f. $t=150\, \log _e\left(\dfrac{59}{40}\right) \ \text {minutes }$

g. $t=50 \ \text{minutes}$

Show Worked Solution

a. $\text{Initial concentration}=\dfrac{5}{3000}=0.001\dot{6} \ \text{kg/L}$

$\text{Concentration entering tank}=0.1 \ \text{kg/L}$

$\text{Since} \ \ 0.1>0.001\dot{6}, \ \text{salt concentration increases with time.}$

♦♦♦ Mean mark (a) 20%.

b. $Q=\text{salt in tank at time}\ t$

$\dfrac{d Q}{d t}=\dfrac{d Q}{d t_{\text {in }}}-\dfrac{d Q}{d t_{\text {out }}}$

$\dfrac{d Q}{d t}=0.1 \times 20-\dfrac{Q}{3000} \times 20=2-\dfrac{Q}{150}=\dfrac{300-Q}{150}$

c. $Q_1=5+15 \times \dfrac{300-5}{150}=34.5 \ \text{kg}$

$Q_2=34.5+15 \times \dfrac{300-34.5}{150}=61.05 \ \text{kg}$

♦ Mean mark (c) 46%.

d.	$\dfrac{d Q}{d t}$	$=\dfrac{300-Q}{150}$
	$\dfrac{d t}{d Q}$	$=\dfrac{150}{300-Q}$
	$\displaystyle \int d t$	$=\displaystyle \int \frac{150}{300-Q} d Q$
	$ t$	$=-150\, \log _e(300-Q)+c$

$\text{When} \ \ t=0, Q=5:$

$0=-150\, \log _e 295+c \ \ \Rightarrow \ \ c=150\, \log _e 295$

$ t$	$=150\, \log _e 295-150\, \log _e(300-Q)$
$ t$	$=150\, \log _e\left(\dfrac{295}{300-Q}\right)$

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

$Q=300-295 e^{-\tfrac{t}{150}}$

e. $\text{As} \ \ t \rightarrow \infty, Q \rightarrow 300$

f. $\text{Find \(t$ when $Q=100$ (using part d):}\)

$t=150\, \log _e\left(\dfrac{295}{300-100}\right)=150\, \log _e\left(\dfrac{59}{40}\right) \ \text {minutes }$

g. $\text{After the tap is turned off:}$

$Q=100+0.1 \times 20 t=100+2 t$

$\text{Volume in tank}=3000+20 t$

$\text{Solve for \(t$:}\)

$\dfrac{1}{20}=\dfrac{100+2 t}{3000+20 t}$

$t=50 \ \text{minutes}$

♦♦♦ Mean mark (g) 18%.

Complex Numbers, SPEC2 2025 VCAA 2

Sketch $\{z: z \bar{z}=4, z \in C\}$ on the Argand plane below. (1 mark)

1. Show that $\{z:|z-2 i|=|z-\sqrt{3}-i|, z \in C\}$ may be expressed as $y=\sqrt{3} x$. (2 marks)
  
  --- 5 WORK AREA LINES (style=lined) ---
2. Sketch $\{z:|z-2 i|=|z-\sqrt{3}-i|, z \in C\}$ on the Argand plane in part a. (1 mark)
  
  --- 0 WORK AREA LINES (style=lined) ---
1. Find the points of intersection of the curves defined in part a and in part b.i, expressing your answers in the form $a+i b$, where $a, b \in R$. (2 marks)
  
  --- 4 WORK AREA LINES (style=lined) ---
2. Label these points on the Argand plane in part a. (1 mark)
  
  --- 0 WORK AREA LINES (style=lined) ---

Consider the points $P$ and $Q$ labelled on the Argand plane below.

A ray originating at point $P$ and passing through point $Q$ has the equation $\operatorname{Arg}\left(z-z_0\right)=\theta$, where $\theta$ is a radian measure.
Write down the values of $z_0$ and $\theta$. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Find the area of the minor segment bounded by the chord connecting the points $P$ and $Q$ and the circle given by $|z|=3$.
Give your answer in the form $c \pi+d$, where $c, d \in R$. (2 marks)

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

Show Answers Only

a. $z \bar{z}=4\ \ \Rightarrow\ \ \text{Circle, centre (0, 0), radius = 2}$

b.i $\abs{z-2 i}=\abs{x+(y-2) i}=x^2+(y-2)^2$

$\abs{z-\sqrt{3}-i}=\abs{x-\sqrt{3}+(y-1)^2}=(x-\sqrt{3})^2+(y-1)^2$

$\text{Equating expressions:}$

$x^2+(y-2)^2=(x-\sqrt{3})^2+(y-1)^2$

$x^2+y^2-4 y+4=x^2-2 \sqrt{3} x+3+y^2-2 y+1$

$-2 y$	$=-2 \sqrt{3} x$
$y$	$=\sqrt{3} x$

b.ii $\text {See image in part (a).}$

$\abs{z-2 i}=\abs{z-\sqrt{3}-i} \ \Rightarrow \ \text{see straight line in image.}$

c.i $\text{Find intersection:}$

$x^2+y^2=4 \ \ \text{and} \ \ y=\sqrt{3} x$

$x^2+3 x^2=4 \ \ \Rightarrow \ \ x^2=1 \ \ \Rightarrow \ \ x= \pm 1$

$\text{Intersection co-ordinates:}\ \ (1, \sqrt{3}),(-1,-\sqrt{3})$

$z=1+\sqrt{3} i$

$z=-1-\sqrt{3} i$

c.ii $\text{See image in part (a)}$

d. $z_0=\dfrac{3}{2}+\dfrac{3 \sqrt{3}}{2} i$

$\theta=-\dfrac{\pi}{3}$

e.	$A$	$=\dfrac{1}{2} \times 3^2 \times\left(\dfrac{\pi}{3}-\sin \dfrac{\pi}{3}\right)$
		$=\dfrac{3}{2} \pi-\dfrac{9 \sqrt{3}}{4}$

Show Worked Solution

a. $z \bar{z}=4\ \ \Rightarrow\ \ \text{Circle, centre (0, 0), radius = 2}$

b.i $\abs{z-2 i}=\abs{x+(y-2) i}=x^2+(y-2)^2$

$\abs{z-\sqrt{3}-i}=\abs{x-\sqrt{3}+(y-1)^2}=(x-\sqrt{3})^2+(y-1)^2$

$\text{Equating expressions:}$

$x^2+(y-2)^2=(x-\sqrt{3})^2+(y-1)^2$

$x^2+y^2-4 y+4=x^2-2 \sqrt{3} x+3+y^2-2 y+1$

$-2 y$	$=-2 \sqrt{3} x$
$y$	$=\sqrt{3} x$

b.ii $\text {See image in part (a).}$

$\abs{z-2 i}=\abs{z-\sqrt{3}-i} \ \Rightarrow \ \text{see straight line in image.}$

c.i $\text{Find intersection:}$

$x^2+y^2=4 \ \ \text{and} \ \ y=\sqrt{3} x$

$x^2+3 x^2=4 \ \ \Rightarrow \ \ x^2=1 \ \ \Rightarrow \ \ x= \pm 1$

$\text{Intersection co-ordinates:}\ \ (1, \sqrt{3}),(-1,-\sqrt{3})$

$z=1+\sqrt{3} i$

$z=-1-\sqrt{3} i$

c.ii $\text{See image in part (a)}$

d. $z_0=\dfrac{3}{2}+\dfrac{3 \sqrt{3}}{2} i$

$\theta=-\dfrac{\pi}{3}$

Mean mark (d) 51%.

e.	$A$	$=\dfrac{1}{2} \times 3^2 \times\left(\dfrac{\pi}{3}-\sin \dfrac{\pi}{3}\right)$
		$=\dfrac{3}{2} \pi-\dfrac{9 \sqrt{3}}{4}$

Calculus, SPEC2 2025 VCAA 1

Sketch the graph of $y(x)=\dfrac{3 x}{x^3+x+2}$ on the axes below.
Label the asymptotes with their equations, and label the turning point and the point of inflection with their coordinates. Give the coordinates of the point of inflection correct to one decimal place. (3 marks)

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

The region bounded by the graph of $y=\dfrac{3 x}{x^3+x+2}$, the coordinate axes and the line $x=2$ is rotated about the $x$-axis to form a solid of revolution.
1. Write down a definite integral that, when evaluated, will give the volume of the solid of revolution. (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---
2. Find the volume of the solid of revolution correct to two decimal places. (1 mark)
  
  --- 2 WORK AREA LINES (style=lined) ---
Find the equations of the vertical asymptotes of the curve given by $y=\dfrac{3 x}{x^3-5 x+2}$. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
A family of curves is given by $y(x)=\dfrac{3 x}{x^3+a x+2}$, where $a \in R$.
1. Consider the case where the graph has a stationary point $P$.
2. Find the $y$-coordinate of $P$ in terms of $a$. (1 mark)
  
  --- 3 WORK AREA LINES (style=lined) ---
3. For a given value of $a$, the graph has no stationary points.
4. Find the equations of the vertical asymptotes of the graph in this case. (1 mark)
  
  --- 3 WORK AREA LINES (style=lined) ---
5. For a given value of $a$, the graph will have a point of inflection at $x=2$.
6. Find the value of $a$. (2 marks)
  
  --- 3 WORK AREA LINES (style=lined) ---

Show Answers Only

b.i. $V=\displaystyle \int_0^2 \pi\left(\frac{3 x}{x^3+x+2}\right)^2 \, d x$

b.ii. $V=2.29\ \text{u}^3$

c. $x=-\sqrt{2}-1, \ x=\sqrt{2}-1, \ x=2$

d.i. $\text{SP at} \ \left(1, \dfrac{3}{3+a}\right)$

d.ii. $\text{No SP’s exist when}\ \ 3+a=0 \ \ \Rightarrow \ \ a=-3$

$x=-2,1$

d.iii. $a=\dfrac{24}{5}$

Show Worked Solution

b.i. $V=\displaystyle \int_0^2 \pi\left(\frac{3 x}{x^3+x+2}\right)^2 \, d x$

b.ii. $V=2.29\ \text{u}^3$

c. $\text{Find vertical asymptotes.}$

$\text {Solve:} \ \ x^3-5 x+\dfrac{1}{2}=0$

$x=-\sqrt{2}-1, \ x=\sqrt{2}-1, \ x=2$

d.i. $y=\dfrac{3 x}{x^3+a x+2}$

$\text{Find \(x$ when $y^{\prime}=0 \ \ \Rightarrow \ \ x=1$}\)

$\text{SP at} \ \ \left(1, \dfrac{3}{3+a}\right)$

d.ii. $\text{No SP’s exist when}\ \ 3+a=0 \ \ \Rightarrow \ \ a=-3$

$y=\dfrac{3 x}{x^3-3 x+2}$

$\text{Vertical asymptotes occur when}\ \ x^3-3 x+2=0$

$\text{(By CAS):} \ \ x=-2,1$

♦ Mean mark (d.ii) 46%.

d.iii. $\text{If POI exists at} \ \ x=2:$

$y^{\prime \prime}(2)=0 \ \Rightarrow \ \dfrac{-3(5 a-24)}{2(a+5)^3}=0$

$a=\dfrac{24}{5}$

Calculus, EXT1 C3 2025 SPEC2 10*

The region bounded by the curve given by $y=3 \cos ^{-1}(x)$, for $0 \leq y \leq a$, where $a>0$, and the line $x=0$ is rotated about the $y$-axis to form a solid of revolution.

If the volume of the solid is $\dfrac{\pi(4 \pi+3 \sqrt{3})}{8}$, determine the value of $a$. (3 marks)

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

Show Answers Only

$a=\pi$

Show Worked Solution

$y=3 \cos ^{-1}(x) \ \Rightarrow \ x=\cos \left(\dfrac{y}{3}\right)$

	$V=\pi \displaystyle \int x^2\, d y$	$=\pi \displaystyle \int_0^a \cos ^2\left(\frac{y}{3}\right)\, d y$
		$=\pi \displaystyle \int_0^a \dfrac{1}{2}\left(1+\cos\dfrac{2y}{3} \right)\,dy$
		$=\pi\,\left[\dfrac{y}{2}+\dfrac{3}{4}\sin \dfrac{2y}{3}\right]_0^a$
		$=\pi\,\left(\dfrac{a}{2}+\dfrac{3}{4}\sin \dfrac{2a}{3}\right)$
		$=\pi\,\left(\dfrac{4a+6 \times \sin\frac{2a}{3}}{8}\right)$

$\text{Equating volumes:}\ \ \dfrac{\pi(4 \pi+3 \sqrt{3})}{8}=\pi\,\left(\dfrac{4a+6 \times \sin\frac{2a}{3}}{8}\right)$

$\Rightarrow a=\pi$

Calculus, EXT1 C2 2025 SPEC2 7 MC

Using the substitution $u=\cos (\theta), \ \dfrac{1}{2} \displaystyle \int_0^{\tfrac{\pi}{2}} \dfrac{\sin (2 \theta)}{1+\cos (\theta)} \, d \theta$ can be expressed as

$\displaystyle\int_0^{\tfrac{\pi}{2}} u \sqrt{\frac{1-u}{1+u}}\ d u$
$\displaystyle\int_0^1\left(1+\frac{1}{1+u}\right) d u$
$\displaystyle\int_0^{\tfrac{\pi}{2}}\left(1-\frac{1}{1+u}\right) d u$
$\displaystyle\int_0^1\left(1-\frac{1}{1+u}\right) d u$

Show Answers Only

$D$

Show Worked Solution

$u=\cos (\theta) \ \Rightarrow \ du=-\sin (\theta)\ d \theta$

$\text{When} \ \ \theta=\dfrac{\pi}{2}, u=0$

$\text{When}\ \ \theta=0, u=1$

$I$	$=\displaystyle\frac{1}{2} \int_0^{\tfrac{\pi}{2}} \frac{\sin (2 \theta)}{1+\cos \theta} \, d \theta$
	$=\displaystyle\int_0^{\tfrac{\pi}{2}} \frac{\sin (\theta) \cos (\theta)}{1+\cos (\theta)} \, d \theta$
	$=\displaystyle \int_1^0-\frac{u}{1+u} \, d u$
	$=\displaystyle\int_0^1 \frac{1+u-1}{1+u} \, d u$
	$=\displaystyle\int_0^1\left(1-\frac{1}{1+u}\right) \, d u$

$\Rightarrow D$

Complex Numbers, EXT2 N1 2025 SPEC2 6*

Let $z \in C$.

Given that $|z|=1$ and $z \neq 1,$ express $\operatorname{Re}\left(\dfrac{1}{1-z}\right)$ in its simplest form. (3 marks)

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

Show Answers Only

$\operatorname{Re}\left(\dfrac{1}{1-z}\right) = \dfrac{1}{2}$

Show Worked Solution

$z=a+b i$

$\abs{z}=a^2+b^2=1$

$\dfrac{1}{1-z}=\dfrac{1}{1-(a+bi)}=\dfrac{1}{(1-a)-bi} \times \dfrac{(1-a)+bi}{(1-a)+bi}=\dfrac{(1-a)+bi}{(1-a)^2+b^2}$

$\operatorname{Re}\left(\dfrac{1}{1-z}\right)$	$=\dfrac{-a+1}{a^2-2 a+b^2+1}$
	$=\dfrac{-a+1}{1-2 a+1}$
	$=\dfrac{-(a-1)}{-2(a-1)}$
	$=\dfrac{1}{2}$

Probability, 2ADV EQ-Bank 12

A new test has been developed for determining whether or not people are carriers of the Gaussian virus.

Two hundred people are tested. A two-way table is being used to record the results.

What is the value of `A`? (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
A person randomly selected from the tested group is a carrier of the virus.
What is the probability that the test results would show this? (1 mark)
--- 4 WORK AREA LINES (style=lined) ---
If the person randomly selected has a negative result from their test, what is the probability they are not a carrier of the virus? (2 marks)

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

Show Answers Only

a. `98`

b. `37/43`

c. `49/55`

Show Worked Solution

a. `A= 200-(74 + 12 + 16)= 98`

b. `P`	`= text(#Positive carriers)/text(Total carriers)`
	`= 74/86`
	`= 37/43`

c. `text(Total number with negative results) = 110`

`text{Total non-carriers with negative results}\ = 98`

`P\text{(not a carrier)|negative result}\ = 98/110=49/55`

Probability, 2ADV EQ-Bank 6 MC

A survey of 370 people was conducted to investigate the association between watching Anime and the age of the person.

The two-way table shows the responses collected.

One response is randomly chosen. If the responder does not watch Anime, what is the probability they are 30 years old or under?

Show Answers Only

$C$

Show Worked Solution

$\text{Total respondents who don’t watch anime}\ = 184$

$\text{Total respondents 30 years and under who don’t watch amine}\ = 61$

$P\text{(30 or under)}|\text{no anime} = \dfrac{61}{184}=0.331… $

$\Rightarrow C$

Probability, 2ADV EQ-Bank 6 MC

A group of 485 people was surveyed. The people were asked whether or not they smoke. The results are recorded in the table.

A person is selected at random from the group.

If the person chosen was a female, what is the chance she is a non-smoker?

Show Answers Only

`D`

Show Worked Solution

`P(text(Female chosen is a non-smoker))`

`= (text(Total female non-smokers))/(text(Total females))`

`= 153/221`

`= 0.6923…`

`=> D`

Probability, 2ADV EQ-Bank 14

Lie detector tests are not always accurate. A lie detector test was administered to 200 people.

The results were:

• 50 people lied. Of these, the test indicated that 40 had lied;
• 150 people did NOT lie. Of these, the test indicated that 20 had lied.

Complete the table using the information above (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
For what percentage of the people tested was the test accurate? (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
What is the probability that the test indicated a lie for a person who did NOT lie? (1 mark)

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

Show Answers Only

a. `text(See Worked Solutions)`

b. `text(85%)`

c. `2/15`

Show Worked Solution

b. `text(Percentage of people with accurate readings)`

`= text(# Accurate readings)/text(Total readings) xx 100`

`= (40+130)/200`

`= 85 text(%)`

c. `text{P(lie detected when NOT a lie)}`

`= 20/150`

`= 2/15`

Statistics, 2ADV EQ-Bank 6 MC

A petrol station records the amount charged to each customer, rounded to the nearest cent. Which statement best describes this random variable?

Discrete, because the charge is rounded and can only take a countable set of values.
Discrete, because the amount charged depends on the number of litres purchased.
Continuous, because the volume of petrol dispensed can take any value within a range.
Continuous, because the amount charged varies between customers.

Show Answers Only

$A$

Show Worked Solution

Although the underlying volume of petrol is continuous, the amount charged is rounded to the nearest cent, meaning it can only take values from a countable set.
This makes the random variable as defined – the charge – discrete.
Option C is the key distractor: it correctly identifies that volume is continuous, but the question defines the variable as the charge.

$\Rightarrow A$

Vectors, SPEC2 2025 VCAA 19 MC

The plane with equation $x+y+z=a$, where $a \in R$, intersects the coordinate axes at three points that form the vertices of a triangle.

The area of this triangle is given by

$\dfrac{a^2 \sqrt{3}}{4}$
$\dfrac{a^2 \sqrt{3}}{2}$
$\dfrac{a^2}{4}$
$a^2 \sqrt{3}$

Show Answers Only

$B$

Show Worked Solution

$\text{Plane:}\ \ x+y+z=a$

$\text{Axis intercepts:}\ \ (a, 0,0),(0, a, 0),(0,0, a)$

$\text{Distance between \((a, 0,0)$ and $(0, a, 0)$ :}\)

$d=\sqrt{(a-0)^2+(0-a)^2+(0-0)^2}=\sqrt{2} a$

$\text{Similarly for the other two sides}$

$\Rightarrow \ \text{Equilateral triangle with side length} \ \sqrt{2} a$

$A=\dfrac{1}{2} a b \sin C=\dfrac{1}{2}(\sqrt{2} a)^2 \times \sin 60^{\circ}=\dfrac{1}{2} \times 2 a^2 \times \dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3} a^2}{2}$

$\Rightarrow B$

Vectors, SPEC2 2025 VCAA 16 MC

The position vector of a particle at time $t$ is given by $\underset{\sim}{ r }(t)=n e^{-2 t} \underset{\sim}{ i }-t^2 \underset{\sim}{ j }$, where $n$ is a positive constant.

For what value of $n$ is the particle's acceleration perpendicular to its velocity when $t=\dfrac{1}{2}$ ?

$2 e$
$\dfrac{e^{0.5}}{2}$
$\dfrac{e}{2}$
$\dfrac{e}{2 \sqrt{2}}$

Show Answers Only

$ C$

Show Worked Solution

$r(t)$	$=n e^{-2 t} \underset{\sim}{i}-t^2 \underset{\sim}{j}$
$v(t)$	$=-2n e^{-2t} \underset{\sim}{i}-2 t\underset{\sim}{i} \ \ \Rightarrow \ \ v\left(\frac{1}{2}\right)=-2 ne^{-1} \underset{\sim}{i}-\underset{\sim}{j}$
$a(t)$	$=4 ne^{-2t} \underset{\sim}{i}-2 \underset{\sim}{j} \ \ \Rightarrow \ \ a\left(\frac{1}{2}\right)=4ne^{-1} \underset{\sim}{i}-2 \underset{\sim}{j}$

$\text{Velocity} \ \perp \ \text{acceleration at}\ \ t=\dfrac{1}{2}:$

$\displaystyle \binom{-\dfrac{2 n}{e}}{-1}\binom{\dfrac{4 n}{e}}{-2}=0$

$-\dfrac{8 n^2}{e^2}+2=0 \ \ \Rightarrow \ \ n^2=\dfrac{e^2}{4} \ \ \Rightarrow \ \ n=\dfrac{e}{2}$

$\Rightarrow C$

Complex Numbers, EXT2 N2 2025 SPEC1 8

Consider the function with rule $f(z)=z^4+6 z^2+25$, where $z \in C$.

Consider $z_1=1+2 i$.
Plot and label $z_1$ and $\overline{z}_1$ on the Argand plane below. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
Given that $1+2 i$ is a solution of $f(z)=0$, find a quadratic factor of $f(z)$. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence, find all remaining solutions of $f(z)=0$. (2 marks)

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

Show Answers Only

a. $z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i$

b. $z^2-2 z+5$

c. $z=-1+2 i, z=-1-2 i$

Show Worked Solution

a. $z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i$

b. $\text{Since} \ \ 1+2 i \ \ \text{is a solution of}\ \ f(z)=0:$

$\Rightarrow 1-2 i \ \ \text{is also a solution (conjugate factor theorem).}$

$\text {Express as a quadratic factor:}$

$(z-(1+2 i))(z-(1-2 i))$	$=((z-1)-2 i)((z-1)+2 i)$
	$=(z-1)^2-4 i^2$
	$=z^2-2 z+5$

c. $f(z)=z^4+6 z^2+25, \quad z \in C$

$\text{Since \(\ z^2-2 z+5\ $ is a factor:}\)

$f(z)$	$=\left(z^2-2 z+5\right)\left(z^2+a z+b\right)$
	$=z^2\left(z^2+a z+b\right)-2 z\left(z^2+a z+b\right)+5\left(z^2+a z+b\right)$
	$=z^4+a z^3+b z^2-2 z^3-2 a z^2-2 b z+5 z^2+5 a z+5 b$
	$=z^4+(a-2) z^3+(b-2 a+5) z^2+(-2 b+5 a) z+5 b$

Mean mark (c) 53%.

$\text{Equating co-efficients:}$

$a-2=0 \ \Rightarrow \ a=2$

$5 b=25 \ \Rightarrow \ b=5$

$f(z)=\left(z^2-2 z+5\right)\left(z^2+2 z+5\right)$

$\text{Solve:} \ \ z^2+2 z+5=0$

$z=\dfrac{-2 \pm \sqrt{2^2-4 \cdot 1 \cdot 5}}{2}=\dfrac{-2 \pm \sqrt{-16}}{2}=-1 \pm 2 i$

$\therefore \ \text{Other solutions:}\ \ z=-1+2 i, z=-1-2 i$

Proof, EXT1 P1 2025 SPEC1 7

Use mathematical induction to prove that

\begin{align*}
\displaystyle \sum_{i=1}^n(i+1)^2=\frac{1}{6} n\left(2 n^2+9 n+13\right) \text { for all integers}\ n\geq 1,
\end{align*}

where $\displaystyle \sum_{i=1}^n(i+1)^2=2^2+3^2+4^2+\ldots+(n+1)^2$. (4 marks)

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

Show Answers Only

$\text{See Worked Solutions}$

Show Worked Solution

$\text{Prove} \ \ \displaystyle \sum_{i=1}^n(i+1)^2=\dfrac{1}{6} n\left(2 n^2+9 n+13\right)$

$\text{If} \ \ n=1:$

$\text{LHS}=(1+1)^2=4$

$\text {RHS}=\dfrac{1}{6} \times 1(2+9+13)=4$

$\therefore \ \text{True for} \ \ n=1$

$\text{Assume true for} \ \ n=k:$

$\text{i.e.} \ \ \displaystyle \sum_{i=1}^k(i+1)^2=\frac{1}{6} k\left(2 k^2+9 k+13\right)\ \ldots\ (1)$

$\text{Prove true for} \ \ n=k+1:$

$\text{i.e.} \ \ \displaystyle \sum_{i=1}^{k+1}(i+1)^2$	$=\dfrac{1}{6}(k+1)\left[2(k+1)^2+9(k+1)+13\right]$
	$=\dfrac{1}{6}(k+1)\left(2 k^2+4 k+2+9 k+9+13\right)$
	$=\dfrac{1}{6}(k+1)\left(2 k^2+13 k+24\right)$

$\operatorname{LHS}$	$=\dfrac{1}{6} k\left(2 k^2+9 k+13\right)+(k+2)^2\ \ \text{(using (1) above)}$
	$=\dfrac{1}{6}\left(2 k^3+9 k^2+13 k\right)+\left(k^2+4 k+4\right)$
	$=\dfrac{1}{6}\left(2 k^3+9 k^2+13 k+6 k^2+24 k+24\right)$
	$=\dfrac{1}{6}\left(2 k^3+15 k^2+37 k+24\right) \quad \text{(see long division below)}$
	$=\dfrac{1}{6}(k+1)\left(2 k^2+13 k+24\right)$
	$=\operatorname{RHS}$

$\Rightarrow \ \text{True for} \ \ n=k+1$

$\therefore \ \text{Since true for \(n=1$, by PMI, true for integers}\ \ n \geq 1.\)

Calculus, EXT1 2025 SPEC1 6

Find the volume of the solid of revolution formed when the area between the curve $y=\sqrt{\dfrac{\arctan (x)}{1+x^2}}$ and the $x$-axis from $x=1$ to $x=\sqrt{3}$ is rotated about the $x$-axis.

Give your answer in exact form. (4 marks)

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

Show Answers Only

$V=\dfrac{7 \pi^3}{288}\ \ \text{u}^3$

Show Worked Solution

$y$	$=\sqrt{\dfrac{\tan ^{-1}(x)}{1+x^2}}$
$V$	$=\displaystyle\pi \int_1^{\sqrt{3}} \frac{\tan ^{-1}(x)}{1+x^2} d x$

$\text{Let} \ \ u=\tan ^{-1}(x) \ \Rightarrow \ du=\dfrac{d x}{1+x^2}$

$\text{When} \ \ x=\sqrt{3}, u=\tan ^{-1} \sqrt{3}=\dfrac{\pi}{3}$

$\text{When} \ \ x=1, u=\tan ^{-1} 1=\dfrac{\pi}{4}$

$V$	$=\displaystyle \pi \int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}} u\,dx$
	$=\dfrac{\pi}{2}\Big[u^2\Big]_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}}$
	$=\dfrac{\pi}{2}\left[\left(\dfrac{\pi}{3}\right)^2-\left(\dfrac{\pi}{4}\right)^2\right]$
	$=\dfrac{\pi^3}{2}\left(\dfrac{1}{9}-\dfrac{1}{16}\right)$
	$=\dfrac{7 \pi^3}{288}\ \ \text{u}^3$

Calculus, EXT2 C1 EQ-Bank 14

Find $\displaystyle \int_0^1\dfrac{3}{2 \log _e 2}\left(\dfrac{1}{(t+1)(2-t)}\right) d t$ (4 marks)

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

Show Answers Only

$I=\dfrac{1}{2}$

Show Worked Solution

$I=\dfrac{3}{2 \ln 2} \displaystyle \int_0^1 \dfrac{t}{(t+1)(2-t)}\, dt$

$\text {Using partial fractions:}$

$\dfrac{t}{(t+1)(2-t)}=\dfrac{A}{(t+1)}+\dfrac{B}{(2-t)}$

$A(2-t)+B(t+1)=t$

$\text{If}\ \ t=2:$

$3 B=2 \ \Rightarrow \ B=\dfrac{2}{3}$

$\text{If}\ \ t=-1:$

$3 A=-1 \ \Rightarrow \ A=-\dfrac{1}{3}$

$I$	$=\displaystyle \frac{3}{2 \ln 2} \int_0^1-\frac{1}{3(t+1)}+\frac{2}{3(2-t)} d t$
	$=\displaystyle\frac{1}{2 \ln 2} \int_0^1-\frac{1}{t+1}+\frac{2}{2-t} d t$
	$=\displaystyle\frac{1}{2 \ln 2}\Big[-\ln \abs{t+1}+2 \ln \abs{2-t}\Big]_0^1$
	$=\displaystyle\frac{1}{2 \ln 2}[(-\ln 2+2 \ln 1)-(2 \ln 1-2 \ln 2)]$
	$=\displaystyle\frac{1}{2 \ln 2}(\ln 2)$
	$=\displaystyle\frac{1}{2}$

Mechanics, EXT2 M1 2025 SPEC1 3

A particle starts from rest at a fixed point $O$ and travels in a straight line.

The velocity, $v$ m s$^{-1}$, of the particle at time $t$ seconds has equation $v(t)=\dfrac{t}{\sqrt{t^2+k}}$, where $k$ is a positive constant and $t \geq 0$.

Use integration to show that the displacement, $x$ metres, of the particle relative to $O$ is given by $x(t)=\sqrt{t^2+k}-\sqrt{k}$. (1 mark)

--- 5 WORK AREA LINES (style=lined) ---
Find the initial acceleration, in terms of $k$, of the particle in m s$^{-2}$. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Another particle starts at $O$ at the same time as the first particle and follows the same path.
Its position relative to $O$ is described by the equation $s(t)=t$.
Three seconds after leaving $O$ the second particle is 1 m ahead of the first particle.
Find the value of $k$. (2 marks)

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

Show Answers Only

a. $v=\dfrac{t}{\sqrt{t^2+k}}$

$x=\displaystyle \int \frac{t}{\sqrt{t^2+k}} d t=\sqrt{t^2+k}+c$

$\text{When} \ \ t=0, x=0:$

$0=\sqrt{k}+c \ \ \Rightarrow \ \ c=-\sqrt{k}$

$\therefore x=\sqrt{t^2+k}-\sqrt{k}$

b. $a=-\sqrt{k}$

c. $k=\dfrac{25}{16}$

Show Worked Solution

a. $v=\dfrac{t}{\sqrt{t^2+k}}$

$x=\displaystyle \int \frac{t}{\sqrt{t^2+k}} d t=\sqrt{t^2+k}+c$

$\text{When} \ \ t=0, x=0:$

$0=\sqrt{k}+c \ \ \Rightarrow \ \ c=-\sqrt{k}$

$\therefore x=\sqrt{t^2+k}-\sqrt{k}$

b. $\text{Find initial acceleration:}$

$v=t\left(t^2+k\right)^{-\tfrac{1}{2}}$

$\text{Using product rule:}$

$\dfrac{dv}{dt}$	$=t\cdot-\dfrac{1}{2} \cdot 2 t\left(t^2+k\right)^{-\tfrac{3}{2}}+\left(t^2+k\right)^{-\tfrac{1}{2}}$
	$=-t^2\left(t^2+k\right)^{-\tfrac{3}{2}}+\left(t^2+k\right)^{-\tfrac{1}{2}}$

$\text{At} \ \ t=0:$

$a=\dfrac{dv}{dt}=0+k^{-\tfrac{1}{2}}=-\sqrt{k}$

c. $\text{1st particle:} \ \ x(3)=\sqrt{9+k}-\sqrt{k}$

$\text{2nd particle:} \ \ s(3)=3$

$\text{Since at \(t=3$, second particle is 1m ahead:}\)

$3-\sqrt{9+k}+\sqrt{k}=1$

$\sqrt{9+k}$	$=2+\sqrt{k}$
$\left(\sqrt{9+k}\right)^2$	$=(2+\sqrt{k})^2$
$9+k$	$=4+4 \sqrt{k}+k$
$4 \sqrt{k}$	$=5$
$\sqrt{k}$	$=\dfrac{5}{4}$
$k$	$=\dfrac{25}{16}$

♦ Mean mark (c) 40%.

Vectors, EXT2 V1 2025 SPEC1 2

Consider the following two lines, $L_1$ and $L_2$.

$L_1$ passes through the point $A_1(2,3,1)$ and has direction $\underset{\sim}{u}=\underset{\sim}{i}+2 \underset{\sim}{j}-\underset{\sim}{k}$.

$L_2$ passes through the point $A_2(1,3,2)$ and has direction $\underset{\sim}{v}=-\underset{\sim}{i}-\underset{\sim}{j}+\underset{\sim}{k}$.

Find the coordinates of the point of intersection of the two lines. (3 marks)

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

Show Answers Only

$\text{Intersection at}\ (3,5,0).$

Show Worked Solution

$\underset{\sim}{u}=\underset{\sim}{i}+2 \underset{\sim}{j}-\underset{\sim}{k} \ \ \text {passes through} \ A_1(2,3,1)$

$L_1:(2+t) \underset{\sim}{i}+(3+2 t) \underset{\sim}{j}+(1-t) \underset{\sim}{k}$

$\underset{\sim}{v}=\underset{\sim}{-i}-\underset{\sim}{j}+\underset{\sim}{k} \ \ \text {passes through}\ A_2(1,3,2)$

$L_2:(1-s) \underset{\sim}{i}+(3-s)\underset{\sim}{j}+(2+s) \underset{\sim}{k}$

$\text{Equating components for intersection:}$

$2+t=1-s \ \ \Rightarrow \ \ s+t=-1\ \ldots\ (1)$

$3+2 t=3-s \ \ \Rightarrow \ \ s+2 t=0\ \ldots\ (2)$

$\text{Subtract: (2) – (1)}$

$t=1 \ \Rightarrow \ s=-2$

$\text{Point of intersection:}$

$(2+1,3+2,1-1) \equiv(3,5,0)$

Calculus, SPEC2 2025 VCAA 11 MC

Given that $y(x)$ is a solution to the differential equation $\dfrac{d y}{d x}=x^2 y^3$, where $y(1)=3$, the domain of $y$ is

$x \leq\left(\dfrac{7}{6}\right)^{\tfrac{1}{3}}$
$x<\left(\dfrac{7}{6}\right)^{\tfrac{1}{3}}$
$x \geq\left(\dfrac{7}{6}\right)^{\tfrac{1}{3}}$
$x>\left(\dfrac{7}{6}\right)^{\tfrac{1}{3}}$

Show Answers Only

$B$

Show Worked Solution

$\text {Solve DE (by CAS):}$

$\dfrac{d y}{d x}=x^2 y^3 \ \ \text{where} \ \ y(1)=3:$

$\dfrac{1}{18}-\dfrac{1}{2 y^2}=\dfrac{x^3}{3}-\dfrac{1}{3} \ \ \Rightarrow \ y= \pm \sqrt{\dfrac{9}{7-6 x^3}}$

$\text{Solve} \ \ \dfrac{9}{7-6 x^3}>0:$

$x<\left(\dfrac{7}{6}\right)^{\tfrac{1}{3}}$

$\Rightarrow B$

Mean mark 55%.

Calculus, SPEC2 2025 VCAA 10 MC

The region bounded by the curve given by $y=3 \cos ^{-1}(x)$, for $0 \leq y \leq a$, where $a>0$, and the line $x=0$ is rotated about the $y$-axis to form a solid of revolution. The volume of the solid is $\dfrac{\pi(4 \pi+3 \sqrt{3})}{8}$.

The value of $a$ is

$\dfrac{\pi}{4}$
$\dfrac{\pi}{3}$
$\dfrac{\pi}{2}$
$\pi$

Show Answers Only

$D$

Show Worked Solution

$y=3 \cos ^{-1}(x) \ \Rightarrow \ x=\cos \left(\dfrac{y}{3}\right)$

$V=\pi \displaystyle \int x^2\, d y=\pi \int \cos ^2\left(\frac{y}{3}\right) d y$

$\text{Solve for} \ a:$

$\pi \displaystyle \int_0^a \cos ^2\left(\frac{y}{3}\right) d y=\frac{\pi(4 \pi+3 \sqrt{3})}{8}$

$a=\pi$

$\Rightarrow D$

Complex Numbers, SPEC2 2025 VCAA 6 MC

Let $z \in C$.

Given that $|z|=1$ and $z \neq 1, \operatorname{Re}\left(\dfrac{1}{1-z}\right)$ is

$-\dfrac{1}{2}$
$0$
$\dfrac{1}{2}$
$\dfrac{\sqrt{3}}{2}$

Show Answers Only

$C$

Show Worked Solution

$z=a+b i$

$\abs{z}=a^2+b^2=1$

$\dfrac{1}{1-z}=\dfrac{1}{1-(a+bi)}=\dfrac{1}{(1-a)-bi} \times \dfrac{(1-a)+bi}{(1-a)+bi}=\dfrac{(1-a)+bi}{(1-a)^2+b^2}$

$\operatorname{Re}\left(\dfrac{1}{1-z}\right)$	$=\dfrac{-a+1}{a^2-2 a+b^2+1}$
	$=\dfrac{-a+1}{1-2 a+1}$
	$=\dfrac{-(a-1)}{-2(a-1)}$
	$=\dfrac{1}{2}$

$\Rightarrow C$

Complex Number, SPEC2 2025 VCAA 5 MC

The equation $z^3+a z^2+b z-52=0$, where $a, b \in R$ and $z \in C$, has a solution $z=2-3 i$.

The value of $a b$ is

$-232$
$-64$
$-8$
$0$

Show Answers Only

$A$

Show Worked Solution

$z^3+a z^2+b z-5 z=0 \ \ \text {has factor}\ \ z=2-3 i$

$(2-3 i)^3+a(2-3 i)^2+b(2-3 i)-52=0$

$\text{Expand by CAS:}$

$(-5 a+2 b-98)+(-12 a-3 b-9) i=0$

$\text {Equating co-efficients:}$

$\text{Solve for \(a, b$ (by CAS) given}\)

$-5 a+2 b-98=0\ \ldots\ (1)$

$-12 a-3 b-9=0\ \ldots\ (2)$

$a=-8, \ b=29 \ \ \Rightarrow \ \ a b=-232$

$\Rightarrow A$

Calculus, SPEC2 2025 VCAA 4 MC

Consider the following algorithm used to estimate a volume of revolution.

The algorithm above will print the value

$5 \pi$
$9 \pi$
$14 \pi$
$29 \pi$

Show Answers Only

$B$

Show Worked Solution

$\text{Working through algorithm:}$

$f(x)=\sqrt{x+1}$

\begin{array}{ccl}
\text { left } & \text { volume } & \text { sum } \\
1 & \pi(f(1))^2 & 2 \pi \\
2 & \pi(f(2))^2 & 2 \pi+3 \pi=5 \pi \\
3 & \pi(f(3))^2 & 5 \pi+4 \pi=9 \pi
\end{array}

$\Rightarrow B$

Graphs, SPEC2 2025 VCAA 3 MC

The graph of $y=\dfrac{x^2+a}{b x+c}$ has an asymptote given by $y=-\dfrac{1}{2} x+\dfrac{1}{4}$ and a $y$-intercept of –2 .

The values of $a, b$ and $c$ are

$a=2, \ b=-2, \ c=-1$
$a=2, \ b=2, \ c=-1$
$a=-2, \ b=-2, \ c=1$
$a=-2, \ b=-2, \ c=-1$

Show Answers Only

$A$

Show Worked Solution

$\text { Expand expression by CAS: }$

$\dfrac{x^2+a}{b x+c}=\dfrac{a b^2+c^2}{b^2(b x+c)}+\dfrac{x}{b}-\dfrac{c}{b^2}$

$\text{Equating coefficients:}$

$\dfrac{x}{b}-\dfrac{c}{b^2}=-\dfrac{1}{2} x-\dfrac{1}{4}\ \ldots\ (1)$

$\text {Since } y(0)=-2$

$\dfrac{a}{c}=-2\ \ldots\ (2)$

$\text{Solve }(1) \text { and }(2) \text { for } a, b, c:$

$a=2, \ b=-2, \ c=-1$

$\Rightarrow A$

Statistics, STD2 S1 2015 HSC 29d*

Data from 200 recent house sales are grouped into class intervals and a cumulative frequency histogram is drawn.

Use the graph to estimate the median house price. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
By completing the table, identify the modal class interval for the house price data. (2 marks)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Class Centre} \rule[-1ex]{0pt}{0pt} & \text{Frequency} \\ \text{(\$'000)} & \\
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \\
\hline
\rule{0pt}{2.5ex} \rule[-1ex]{0pt}{0pt} & \\
\hline
\end{array}

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

Show Answers Only

`text(From the graph, the estimated median house price = $392 500)`

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Class Centre} \rule[-1ex]{0pt}{0pt} & \text{Frequency} \\ \text{(\$’000)} & \\
\hline
\rule{0pt}{2.5ex} 375 \rule[-1ex]{0pt}{0pt} & 30\\
\hline
\rule{0pt}{2.5ex} 385 \rule[-1ex]{0pt}{0pt} & 50 \\
\hline
\rule{0pt}{2.5ex} 395 \rule[-1ex]{0pt}{0pt} & 70 \\
\hline
\rule{0pt}{2.5ex} 405 \rule[-1ex]{0pt}{0pt} & 50 \\
\hline
\end{array}

`text{Modal class interval}\ = $390\ 000-$400\ 000\ (\text{or}\ $395\ 000)`

Show Worked Solution

`text(From the graph, the estimated median house price = $392 500)`

`text{Modal class interval}\ = $390\ 000-$400\ 000\ (\text{or}\ $395\ 000)`

Statistics, STD2 S1 2010 HSC 26b*

A new shopping centre has opened near a primary school. A survey is conducted to determine the number of motor vehicles that pass the school each afternoon between 2.30 pm and 4.00 pm.

The results for 60 days have been recorded in the table and are displayed in the cumulative frequency histogram.

Find the value of Χ in the table. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
On the cumulative frequency histogram above, draw a cumulative frequency polygon (ogive) for this data. (1 mark)
Use your graph to determine the median. Show, by drawing lines on your graph, how you arrived at your answer. (1 mark)

Show Answers Only

a. `15`

b. & c.

`text(Median)\ ~~155`

Show Worked Solution

a. `X= 25-10= 15`

b. & c.

`text(Median)\ ~~155`

Graphs, SPEC1 2025 VCAA 9

Let $f: R \backslash\{-1,1\} \rightarrow R, f(x)=\dfrac{x^3+x^2-2 x}{1-x^2}$.

Show that $f(x)$ can be written in the form $f(x)=-x-1+\dfrac{1}{x+1}$, for $x \in R \backslash\{-1,1\}$. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
Consider the function with rule
1. \begin{align*}
  g(x)=\left\{\begin{array}{cl}
  \dfrac{x^3+x^2-2 x}{1-x^2}, & x \in R \backslash\{-1,1\} \\
  k, & x \in\{1\}
  \end{array}\right.
  \end{align*}
Find the value of $k$ such that the graph of $g$ is continuous at $x=1$. (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
Sketch the graph of $y=f(x)$ on the axes below.
Label the asymptotes with their equations. (3 marks)

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

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

Show Answers Only

a.	$f(x)$	$=\dfrac{x^3+x^2-2 x}{1-x^2}$
		$=\dfrac{x\left(x^2+x-2\right)}{(1-x)(1+x)}$
		$=\dfrac{-x(x+2)(1-x)}{(1-x)(1+x)}$
		$=\dfrac{-x(x+2)}{x+1}$
		$=\dfrac{-x^2-2 x}{x+1}$
		$=\dfrac{-(x+1)^2+1}{x+1}$
		$=-x-1+\dfrac{1}{x+1}$

b. $k=-\dfrac{3}{2}$

Show Worked Solution

a.	$f(x)$	$=\dfrac{x^3+x^2-2 x}{1-x^2}$
		$=\dfrac{x\left(x^2+x-2\right)}{(1-x)(1+x)}$
		$=\dfrac{-x(x+2)(1-x)}{(1-x)(1+x)}$
		$=\dfrac{-x(x+2)}{x+1}$
		$=\dfrac{-x^2-2 x}{x+1}$
		$=\dfrac{-(x+1)^2+1}{x+1}$
		$=-x-1+\dfrac{1}{x+1}$

b. $\text{Find \(k$ such that $g(x)$ is continuous:}\)

$k$	$=\dfrac{k^3+k^2-2 k}{1-k^2}$
$k-k^3$	$=k^3+k^2-2 k$
$1-k^2$	$=k^2+k-2$
$0$	$=2 k^2+k-3$
$0$	$=(2 k+3)(k-1)$

$\therefore k=-\dfrac{3}{2}\ \ (k \neq 1)$

♦ Mean mark (b) 46%.

c. $\text{Intercepts where}\ \ -x-1+\dfrac{1}{x+1}=0:$

$(x+1)^2=1 \ \ \Rightarrow \ \ x^2+2 x=0 \ \ \Rightarrow \ \ x=0 \ \ \text{or} \ -2$

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

$\text{Hole at} \ \left(1,-\dfrac{3}{2}\right)$

♦♦ Mean mark (c) 37%.

Complex Numbers, SPEC1 2025 VCAA 8

Consider the function with rule $f(z)=z^4+6 z^2+25$, where $z \in C$.

Consider $z_1=1+2 i$.
Plot and label $z_1$ and $\overline{z}_1$ on the Argand plane below. (1 mark)

--- 0 WORK AREA LINES (style=lined) ---
Given that $1+2 i$ is a solution of $f(z)=0$, find a quadratic factor of $f(z)$. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---
Hence, find all remaining solutions of $f(z)=0$. (2 marks)

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

Show Answers Only

a. $z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i$

b. $z^2-2 z+5$

c. $z=-1+2 i, z=-1-2 i$

Show Worked Solution

a. $z_1=1+z_i \ \Rightarrow \ \overline{z}_1=1-z_i$

b. $\text{Since} \ \ 1+2 i \ \ \text{is a solution of}\ \ f(z)=0:$

$\Rightarrow 1-2 i \ \ \text{is also a solution.}$

$\text {Express as a quadratic factor:}$

$(z-(1+2 i))(z-(1-2 i))$	$=((z-1)-2 i)((z-1)+2 i)$
	$=(z-1)^2-4 i^2$
	$=z^2-2 z+5$

c. $f(z)=z^4+6 z^2+25, \quad z \in C$

$\text{Since \(\ z^2-2 z+5\ $ is a factor:}\)

$f(z)$	$=\left(z^2-2 z+5\right)\left(z^2+a z+b\right)$
	$=z^2\left(z^2+a z+b\right)-2 z\left(z^2+a z+b\right)+5\left(z^2+a z+b\right)$
	$=z^4+a z^3+b z^2-2 z^3-2 a z^2-2 b z+5 z^2+5 a z+5 b$
	$=z^4+(a-2) z^3+(b-2 a+5) z^2+(-2 b+5 a) z+5 b$

Mean mark (c) 53%.

$\text{Equating co-efficients:}$

$a-2=0 \ \Rightarrow \ a=2$

$5 b=25 \ \Rightarrow \ b=5$

$f(z)=\left(z^2-2 z+5\right)\left(z^2+2 z+5\right)$

$\text{Solve:} \ \ z^2+2 z+5=0$

$z=\dfrac{-2 \pm \sqrt{2^2-4 \cdot 1 \cdot 5}}{2}=\dfrac{-2 \pm \sqrt{-16}}{2}=-1 \pm 2 i$

$\therefore \ \text{Other solutions:}\ \ z=-1+2 i, z=-1-2 i$

Proof, SPEC1 2025 VCAA 7

Use mathematical induction to prove that

\begin{align*}
\displaystyle \sum_{i=1}^n(i+1)^2=\frac{1}{6} n\left(2 n^2+9 n+13\right) \text { for } n \in N \text {, }
\end{align*}

where $\displaystyle \sum_{i=1}^n(i+1)^2=2^2+3^2+4^2+\ldots+(n+1)^2$. (4 marks)

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

Show Answers Only

$\text{See Worked Solutions}$

Show Worked Solution

$\text{Prove} \ \ \displaystyle \sum_{i=1}^n(i+1)^2=\dfrac{1}{6} n\left(2 n^2+9 n+13\right)$

$\text{If} \ \ n=1:$

$\text{LHS}=(1+1)^2=4$

$\text {RHS}=\dfrac{1}{6} \times 1(2+9+13)=4$

$\therefore \ \text{True for} \ \ n=1$

$\text{Assume true for} \ \ n=k:$

$\text{i.e.} \ \ \displaystyle \sum_{i=1}^k(i+1)^2=\frac{1}{6} k\left(2 k^2+9 k+13\right)\ \ldots\ (1)$

$\text{Prove true for} \ \ n=k+1:$

$\text{i.e.} \ \ \displaystyle \sum_{i=1}^{k+1}(i+1)^2$	$=\dfrac{1}{6}(k+1)\left[2(k+1)^2+9(k+1)+13\right]$
	$=\dfrac{1}{6}(k+1)\left(2 k^2+4 k+2+9 k+9+13\right)$
	$=\dfrac{1}{6}(k+1)\left(2 k^2+13 k+24\right)$

$\operatorname{LHS}$	$=\dfrac{1}{6} k\left(2 k^2+9 k+13\right)+(k+2)^2\ \ \text{(using (1) above)}$
	$=\dfrac{1}{6}\left(2 k^3+9 k^2+13 k\right)+\left(k^2+4 k+4\right)$
	$=\dfrac{1}{6}\left(2 k^3+9 k^2+13 k+6 k^2+24 k+24\right)$
	$=\dfrac{1}{6}\left(2 k^3+15 k^2+37 k+24\right) \quad \text{(see long division below)}$
	$=\dfrac{1}{6}(k+1)\left(2 k^2+13 k+24\right)$
	$=\operatorname{RHS}$

$\Rightarrow \ \text{True for} \ \ n=k+1$

$\therefore \ \text{Since true for \(n=1$, by PMI, true for $\ n \in N$.}\)

Calculus, SPEC1 2025 VCAA 6

Give your answer in the form $\dfrac{a \pi^b}{c}$, where $a, b, c \in Z^{+}$. (4 marks)

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

Show Answers Only

$V=\dfrac{7 \pi^3}{288}\ \text{u}^3$

Show Worked Solution

$y$	$=\sqrt{\dfrac{\tan ^{-1}(x)}{1+x^2}}$
$V$	$=\displaystyle\pi \int_1^{\sqrt{3}} \frac{\tan ^{-1}(x)}{1+x^2} d x$

$\text{Let} \ \ u=\tan ^{-1}(x) \ \Rightarrow \ du=\dfrac{d x}{1+x^2}$

$\text{When} \ \ x=\sqrt{3}, u=\tan ^{-1} \sqrt{3}=\dfrac{\pi}{3}$

$\text{When} \ \ x=1, u=\tan ^{-1} 1=\dfrac{\pi}{4}$

$V$	$=\displaystyle \pi \int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}} u\,dx$
	$=\dfrac{\pi}{2}\Big[u^2\Big]_{\tfrac{\pi}{4}}^{\tfrac{\pi}{3}}$
	$=\dfrac{\pi}{2}\left[\left(\dfrac{\pi}{3}\right)^2-\left(\dfrac{\pi}{4}\right)^2\right]$
	$=\dfrac{\pi^3}{2}\left(\dfrac{1}{9}-\dfrac{1}{16}\right)$
	$=\dfrac{\pi^3}{2}\left(\dfrac{16-9}{9 \times 16}\right)$
	$=\dfrac{7 \pi^3}{288}\ \text{u}^3$

Calculus, SPEC1 2025 VCAA 4

The waiting time, $T$ hours, to see a particular doctor at a clinic has a distribution with a probability density function $f$ defined by

\begin{align*}
f(t)=\left\{\begin{array}{cl}
\dfrac{3}{2 \log _e(2)}\left(\dfrac{1}{(t+1)(2-t)}\right), & 0<t \leq 1 \\
0, & \text{elsewhere}
\end{array}\right.
\end{align*}

Use integration to show that $E (T)=\dfrac{1}{2}$. (3 marks)

--- 10 WORK AREA LINES (style=lined) ---
For random samples of 25 waiting times, it may be assumed that the sample means are approximately normally distributed.
Find the probability that the average waiting time for a random sample of 25 patients is between 0.44 hours and 0.5 hours.
Use $\sigma=0.3$ and $\operatorname{Pr}(Z<1)=0.84$ (2 marks)

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

Show Answers Only

a. $\text{See Worked Solutions}$

b. $\operatorname{Pr}(0.44<\overline{T}<0.5)=0.34$

Show Worked Solution

a. $E(T)=\dfrac{3}{2 \ln 2} \displaystyle \int_0^1 \dfrac{t}{(t+1)(2-t)}\, dt$

$\text {Using partial fractions:}$

$\dfrac{t}{(t+1)(2-t)}=\dfrac{A}{(t+1)}+\dfrac{B}{(2-t)}$

$A(2-t)+B(t+1)=t$

$\text{If}\ \ t=2:$

$3 B=2 \ \Rightarrow \ B=\dfrac{2}{3}$

$\text{If}\ \ t=-1:$

$3 A=-1 \ \Rightarrow \ A=-\dfrac{1}{3}$

Mean mark (a) 51%.

$E(T)$	$=\displaystyle \frac{3}{2 \ln 2} \int_0^1-\frac{1}{3(t+1)}+\frac{2}{3(2-t)} d t$
	$=\displaystyle\frac{1}{2 \ln 2} \int_0^1-\frac{1}{t+1}+\frac{2}{2-t} d t$
	$=\displaystyle\frac{1}{2 \ln 2}\Big[-\ln \abs{t+1}+2 \ln \abs{2-t}\Big]_0^1$
	$=\displaystyle\frac{1}{2 \ln 2}[(-\ln 2+2 \ln 1)-(2 \ln 1-2 \ln 2)]$
	$=\displaystyle\frac{1}{2 \ln 2}(\ln 2)$
	$=\displaystyle\frac{1}{2}$

b. $E(\overline{T})=\dfrac{1}{2}, \ \operatorname{sd}(\overline{T})=\dfrac{\sigma}{\sqrt{n}}=\dfrac{0.3}{\sqrt{25}}=0.06$

$\operatorname{Pr}(0.44<\overline{T}<0.5)$	$=\operatorname{Pr}\left(\dfrac{0.44-0.5}{0.06}<Z<0\right)$
	$=\operatorname{Pr}(-1<Z<0)$
	$=\operatorname{Pr}(0<Z<1)$
	$=0.84-0.50$
	$=0.34$

Calculus, SPEC1 2025 VCAA 3

A particle starts from rest at a fixed point $O$ and travels in a straight line.

The velocity, $v$ m s$^{-1}$, of the particle at time $t$ seconds has equation $v(t)=\dfrac{t}{\sqrt{t^2+k}}$, where $k$ is a positive constant and $t \geq 0$.

Use integration to show that the displacement, $x$ metres, of the particle relative to $O$ is given by $x(t)=\sqrt{t^2+k}-\sqrt{k}$. (1 mark)

--- 5 WORK AREA LINES (style=lined) ---
Find the initial acceleration, in terms of $k$, of the particle in m s$^{-2}$. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Another particle starts at $O$ at the same time as the first particle and follows the same path.
Its position relative to $O$ is described by the equation $s(t)=t$.
Three seconds after leaving $O$ the second particle is 1 m ahead of the first particle.
Find the value of $k$. (2 marks)

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

Show Answers Only

a. $v=\dfrac{t}{\sqrt{t^2+k}}$

$x=\displaystyle \int \frac{t}{\sqrt{t^2+k}} d t=\sqrt{t^2+k}+c$

$\text{When} \ \ t=0, x=0:$

$0=\sqrt{k}+c \ \ \Rightarrow \ \ c=-\sqrt{k}$

$\therefore x=\sqrt{t^2+k}-\sqrt{k}$

b. $a=-\sqrt{k}$

c. $k=\dfrac{25}{16}$

Show Worked Solution

a. $v=\dfrac{t}{\sqrt{t^2+k}}$

$x=\displaystyle \int \frac{t}{\sqrt{t^2+k}} d t=\sqrt{t^2+k}+c$

$\text{When} \ \ t=0, x=0:$

$0=\sqrt{k}+c \ \ \Rightarrow \ \ c=-\sqrt{k}$

$\therefore x=\sqrt{t^2+k}-\sqrt{k}$

b. $\text{Find initial acceleration:}$

$v=t\left(t^2+k\right)^{-\tfrac{1}{2}}$

$\text{Using product rule:}$

$\dfrac{dv}{dt}$	$=t\cdot-\dfrac{1}{2} \cdot 2 t\left(t^2+k\right)^{-\tfrac{3}{2}}+\left(t^2+k\right)^{-\tfrac{1}{2}}$
	$=-t^2\left(t^2+k\right)^{-\tfrac{3}{2}}+\left(t^2+k\right)^{-\tfrac{1}{2}}$

$\text{At} \ \ t=0:$

$a=\dfrac{dv}{dt}=0+k^{-\tfrac{1}{2}}=-\sqrt{k}$

c. $\text{1st particle:} \ \ x(3)=\sqrt{9+k}-\sqrt{k}$

$\text{2nd particle:} \ \ s(3)=3$

$\text{Since at \(t=3$, second particle is 1m ahead:}\)

$3-\sqrt{9+k}+\sqrt{k}=1$

$\sqrt{9+k}$	$=2+\sqrt{k}$
$\left(\sqrt{9+k}\right)^2$	$=(2+\sqrt{k})^2$
$9+k$	$=4+4 \sqrt{k}+k$
$4 \sqrt{k}$	$=5$
$\sqrt{k}$	$=\dfrac{5}{4}$
$k$	$=\dfrac{25}{16}$

♦ Mean mark (c) 40%.

Vectors, SPEC1 2025 VCAA 2

Consider the following two lines, $L_1$ and $L_2$.

$L_1$ passes through the point $A_1(2,3,1)$ and has direction $\underset{\sim}{u}=\underset{\sim}{i}+2 \underset{\sim}{j}-\underset{\sim}{k}$.

$L_2$ passes through the point $A_2(1,3,2)$ and has direction $\underset{\sim}{v}=-\underset{\sim}{i}-\underset{\sim}{j}+\underset{\sim}{k}$.

Find the coordinates of the point of intersection of the two lines. (3 marks)

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

Show Answers Only

$\text{Intersection at}\ (3,5,0).$

Show Worked Solution

$\underset{\sim}{u}=\underset{\sim}{i}+2 \underset{\sim}{j}-\underset{\sim}{k} \ \ \text {passes through} \ A_1(2,3,1)$

$L_1:(2+t) \underset{\sim}{i}+(3+2 t) \underset{\sim}{j}+(1-t) \underset{\sim}{k}$

$\underset{\sim}{v}=\underset{\sim}{-i}-\underset{\sim}{j}+\underset{\sim}{k} \ \ \text {passes through}\ A_2(1,3,2)$

$L_2:(1-s) \underset{\sim}{i}+(3-s)\underset{\sim}{j}+(2+s) \underset{\sim}{k}$

$\text{Equating components for intersection:}$

$2+t=1-s \ \ \Rightarrow \ \ s+t=-1\ \ldots\ (1)$

$3+2 t=3-s \ \ \Rightarrow \ \ s+2 t=0\ \ldots\ (2)$

$\text{Subtract: (2) – (1)}$

$t=1 \ \Rightarrow \ s=-2$

$\text{Point of intersection:}$

$(2+1,3+2,1-1) \equiv(3,5,0)$

Calculus, SPEC1 2025 VCAA 1

Consider the curve with equation $x e^{-2 y}+y^2 e^x=8 e^4$.

Find the equation of the tangent to the curve at the point $(4,-2)$. (4 marks)

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

Show Answers Only

$y=\dfrac{5}{12} x-\dfrac{11}{3}$

Show Worked Solution

$x e^{-2 y}+y^2 e^x=8 e^4$

$x\left(-2 e^{-2 y}\right) \dfrac{d y}{d x}+e^{-2 y}+2 y e^x \dfrac{d y}{d x}+y^2 e^x=0$

$\dfrac{d y}{d x}\left(-2 x e^{-2 y}+2 y e^x\right)=-e^{-2 y}-y^2 e^x$

$\dfrac{d y}{d x}=\dfrac{-e^{-2 y}-y^2 e^x}{-2 x e^{-2 y}+2 y e^x}$

$\text{Find \(\ \dfrac{d y}{d x}\ $ at $\ (4,-2)$:}\)

$\dfrac{d y}{d x}=\dfrac{-e^4-4 e^4}{-8 e^4-4 e^4}=\dfrac{-5 e^4}{-12 e^4}=\dfrac{5}{12}$

$\text{Equation of tangent, \(m=\dfrac{5}{12} $ through $(4,-2)$:}\)

$y+2$	$=\dfrac{5}{12}(x-4)$
$y+2$	$=\dfrac{5}{12} x-\dfrac{5}{3}$
$y$	$=\dfrac{5}{12} x-\dfrac{11}{3}$

Trigonometry, 2ADV T3 2025 MET1 3

Let $f(x)=2 \cos (2 x)+1$ over the domain $x \in\left[0, 2 \pi \right]$.

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

--- 2 WORK AREA LINES (style=lined) ---
Solve $f(x)=0$ for $x$. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Sketch the graph of $y=f(x)$ for $x \in\left[\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\right]$ on the axes below.
Label the endpoints with their coordinates. (2 marks)

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

Show Answers Only

a. $\text{Range of } f(x):-1 \leqslant y \leqslant 3$

b. $x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}$

Show Worked Solution

a. $\text{Amplitude}=2 \ \ \text{about} \ \ y=1.$

$\text{Range of } f(x):\ -1 \leqslant y \leqslant 3$

b.	$2 \cos (2 x)+1$	$=0$
	$\cos (2 x)$	$=-\dfrac{1}{2}$

$\text{Base angle}=\dfrac{\pi}{3}$

$2x=\pi-\dfrac{\pi}{3}, \pi+\dfrac{\pi}{3}, \cdots$

$2x=\dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{8 \pi}{3}, \dfrac{10 \pi}{3}$

$x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}$

♦ Mean mark (c) 49%.

Probability, MET1 2025 VCAA 6

Consider the binomial random variable $X \sim \operatorname{Bi}\left(6, \dfrac{1}{4}\right)$.

Find $\operatorname{var}(X)$. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Determine $\operatorname{Pr}(X \geq 5)$.
Give your answer in the form $\dfrac{a}{2^b}$, where $a, b \in Z$. (2 marks)

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

Show Answers Only

a. $\dfrac{9}{8}$

b. $\dfrac{19}{2^{12}}$

Show Worked Solution

a. $X \sim \operatorname{Bi}\left(6, \dfrac{1}{4}\right)$

$\operatorname{var}(X)=n p(1-p)=6 \times \dfrac{1}{4} \times \dfrac{3}{4}=\dfrac{18}{16}=\dfrac{9}{8}$

b.	$\operatorname{Pr}$	$(X \geqslant 5)$
		$=\operatorname{Pr}(X=5)+\operatorname{Pr}(X=6)$
		$={ }^6 C_5\left(\dfrac{1}{4}\right)^5\left(\dfrac{3}{4}\right)^1+{ }^6 C_6\left(\dfrac{1}{4}\right)^6\left(\dfrac{3}{4}\right)^0$
		$=6 \times \dfrac{3}{4^6}+\dfrac{1}{4^6}$
		$=\dfrac{19}{4^6}$
		$=\dfrac{19}{2^{12}}$

Mean mark (b) 52%.

L&E, 2ADV E1 2025 MET1 5

Solve $e^{2 x}-8 e^x+7=0$ for $x$. (3 marks)

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

Show Answers Only

$x=\ln 7, \ x=0$

Show Worked Solution

$\text{Solve}\ \ e^{2 x}-8 e^x+7=0:$

$\text{Let} \ \ X=e^x$

$X^2-8 X+7$	$=0$
$(X-7)(X-1)$	$=0$

$x=7 \ \text{or 1}$

$e^x=7 \ \ \qquad e^x=1$

$x=\ln 7 \qquad x=0$

Calculus, MET1 2025 VCAA 5

Solve $e^{2 x}-8 e^x+7=0$ for $x$. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
Let $g(x)=e^{2 x}-8 e^x+7$, where $x \in R$.
The function $g(x)$ has exactly one stationary point, a local minimum.
Find the largest value of $a$ such that when $g$ is restricted to the domain $(-\infty, a]$ it has an inverse function. (2 marks)

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

Show Answers Only

a. $x=\ln 7 \qquad x=0$

b. $(-\infty, \ln 4]$

Show Worked Solution

a. $\text{Solve}\ \ e^{2 x}-8 e^x+7=0:$

$\text{Let} \ \ X=e^x$

$X^2-8 X+7$	$=0$
$(X-7)(X-1)$	$=0$

$x=7 \ \text{or 1}$

$e^x=7 \ \ \qquad e^x=1$

$x=\ln 7 \qquad x=0$

b. $g(x)=e^{2 x}-8 e^x+7$

$g^{\prime}(x)=2 e^{2 x}-8 e^x=2 e^x\left(e^x-4\right)$

$\text{SP occurs when}\ \ 2 e^x\left(e^x-4\right)=0:$

$e^{x} \neq 0 $

$e^x-4=0 \ \ \Rightarrow \ \ e^x=4 \ \ \Rightarrow \ \ x=\ln 4$

$\therefore(-\infty, \ln 4] \ \text{is the largest domain where \(g(x)$ has an inverse.}\)

Mean mark (b) 51%.

Probability, 2ADV S1 2025 MET1 4

The probability distribution for the discrete random variable $X$ is given in the table below, where $k$ is a positive real number.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}x \rule[-1ex]{0pt}{0pt}& \quad \quad 0 \quad \quad & \quad \quad 1 \quad \quad & \quad \quad 2 \quad \quad & \quad \quad 3 \quad \quad \\
\hline
\rule{0pt}{2.5ex}\operatorname{Pr}(X=x) \rule[-1ex]{0pt}{0pt}& \dfrac{4}{k} &\dfrac{2 k}{75} &\dfrac{k}{75} & \dfrac{2}{k} \\
\hline
\end{array}

Show that $k=10$ or $k=15$. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
Let $k=15$.
1. Find $\operatorname{Pr}(X>1)$. (1 mark)
  
  --- 3 WORK AREA LINES (style=lined) ---
2. Find $E (X)$. (1 mark)
  
  --- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $\text{Sum of probabilities\(=1$:}\)

$\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}$	$=1$
$\dfrac{6}{k}+\dfrac{3 k}{75}$	$=1$
$3 k^2+450$	$=75k$
$3 k^2-75 k+450$	$=0$
$k^2-25 k+150$	$=0$
$(k-10)(k-15)$	$=0$

$\therefore k=10 \ \ \text{or 15}$

b.i. $\operatorname{Pr}(X>1)=\dfrac{1}{3}$

b.ii. $E(X)=\dfrac{90}{75}$

Show Worked Solution

a. $\text{Sum of probabilities\(=1$:}\)

$\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}$	$=1$
$\dfrac{6}{k}+\dfrac{3 k}{75}$	$=1$
$3 k^2+450$	$=75k$
$3 k^2-75 k+450$	$=0$
$k^2-25 k+150$	$=0$
$(k-10)(k-15)$	$=0$

$\therefore k=10 \ \ \text{or 15}$

b.i.	$\operatorname{Pr}(X>1)$	$=\operatorname{Pr}(X=2)+\operatorname{Pr}(X=3)$
		$=\dfrac{15}{75}+\dfrac{2}{15}$
		$=\dfrac{1}{3}$

b.ii.	$E(X)$	$=0 \times \dfrac{4}{15}+1 \times \dfrac{30}{75}+2 \times \dfrac{15}{75}+3 \times \dfrac{2}{15}$
		$=\dfrac{6}{5}$

Probability, MET1 2025 VCAA 4

The probability distribution for the discrete random variable $X$ is given in the table below, where $k$ is a positive real number.

Show that $k=10$ or $k=15$. (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
Let $k=15$.
1. Find $\operatorname{Pr}(X>1)$. (1 mark)
  
  --- 3 WORK AREA LINES (style=lined) ---
2. Find $E (X)$. (1 mark)
  
  --- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a. $\text{Sum of probabilities\(=1$:}\)

$\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}$	$=1$
$\dfrac{6}{k}+\dfrac{3 k}{75}$	$=1$
$3 k^2+450$	$=75k$
$3 k^2-75 k+450$	$=0$
$k^2-25 k+150$	$=0$
$(k-10)(k-15)$	$=0$

$\therefore k=10 \ \ \text{or 15}$

b.i. $\operatorname{Pr}(X>1)=\dfrac{1}{3}$

b.ii. $E(X)=\dfrac{90}{75}$

Show Worked Solution

a. $\text{Sum of probabilities\(=1$:}\)

$\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}$	$=1$
$\dfrac{6}{k}+\dfrac{3 k}{75}$	$=1$
$3 k^2+450$	$=75k$
$3 k^2-75 k+450$	$=0$
$k^2-25 k+150$	$=0$
$(k-10)(k-15)$	$=0$

$\therefore k=10 \ \ \text{or 15}$

b.i.	$\operatorname{Pr}(X>1)$	$=\operatorname{Pr}(X=2)+\operatorname{Pr}(X=3)$
		$=\dfrac{15}{75}+\dfrac{2}{15}$
		$=\dfrac{1}{3}$

b.ii.	$E(X)$	$=0 \times \dfrac{4}{15}+1 \times \dfrac{30}{75}+2 \times \dfrac{15}{75}+3 \times \dfrac{2}{15}$
		$=\dfrac{30}{75}+\dfrac{30}{75}+\dfrac{30}{75}$
		$=\dfrac{90}{75}$

BIOLOGY, M8 EQ-Bank 26

The biological structure shown is part of one of the systems in the body.

Name the biological structure. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
Outline how neural pathways allow the hypothalamus to maintain body temperature within normal range. (3 marks)

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

Show Answers Only

a. Nerve cell/neuron

b. Steps involved in system’s response to a stimulus:

Thermoreceptors detect a change in body temperature and send nerve impulses to the hypothalamus.
The hypothalamus (control centre) processes the signal and sends neural signals to effectors.
Effectors respond (e.g. sweat glands, skeletal muscles, blood vessels) to restore temperature to the normal range.

Show Worked Solution

a. Nerve cell/neuron

b. Steps involved in system’s response to a stimulus:

Thermoreceptors detect a change in body temperature and send nerve impulses to the hypothalamus.
The hypothalamus (control centre) processes the signal and sends neural signals to effectors.
Effectors respond (e.g. sweat glands, skeletal muscles, blood vessels) to restore temperature to the normal range.

Graphs, MET1 2025 VCAA 3

Let $f:[0,2 \pi] \rightarrow R, f(x)=2 \cos (2 x)+1$.

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

--- 2 WORK AREA LINES (style=lined) ---
Solve $f(x)=0$ for $x$. (3 marks)

--- 8 WORK AREA LINES (style=lined) ---
Sketch the graph of $y=f(x)$ for $x \in\left[\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\right]$ on the axes below.
Label the endpoints with their coordinates. (2 marks)

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

Show Answers Only

a. $\text{Range of } f(x):-1 \leqslant y \leqslant 3$

b. $x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}$

Show Worked Solution

a. $\text{Amplitude}=2 \ \ \text{about} \ \ y=1.$

$\text{Range of } f(x):\ -1 \leqslant y \leqslant 3$

b.	$2 \cos (2 x)+1$	$=0$
	$\cos (2 x)$	$=-\dfrac{1}{2}$

$\text{Base angle}=\dfrac{\pi}{3}$

$2x=\pi-\dfrac{\pi}{3}, \pi+\dfrac{\pi}{3}, \cdots$

$2x=\dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{8 \pi}{3}, \dfrac{10 \pi}{3}$

$x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}$

♦ Mean mark (c) 49%.

\(d\)	\(=\sqrt{(x-2)^2+(x+c)^2}\)
	\(=\sqrt{x^2-4x+4+x^2+2cx+c^2}\)
	\(=\sqrt{2x^2+(2c-4)x+4+c^2}\)

\(W\)	\(= F \cos\theta \times d\)
	\(= 600 \cos 60^{\circ} \times 10\)
	\(= 600 \times 0.5 \times 10\)
	\(= 3000 \ \text{J}\)

a.i.	\(\mu\)	\(=1000\)
	\(\sigma\)	\(=\dfrac{80}{\sqrt{16}}=16\)

\(1\)	\(\geqslant 1.96 \times \dfrac{5}{\sqrt{n}}\)
\(n\)	\(\geqslant 96.04\)
\(n\)	\(=97\)

f.i.	\(p\)	\(=\operatorname{Pr}\left(z<\dfrac{748-750}{\frac{5}{\sqrt{50}}}\right)\)
		\(=\operatorname{Pr}\left(z<-\dfrac{2 \sqrt{50}}{5}\right)\)
		\(=0.0023\)

\(\text{95% CI}\)	\(=\left(750-1.96 \times \dfrac{5}{\sqrt{30}}, 750+1.96\times\dfrac{5}{\sqrt{30}}\right)\)
	\(=(748.2,751.8)\)

d.	\(\dfrac{d Q}{d t}\)	\(=\dfrac{300-Q}{150}\)
	\(\dfrac{d t}{d Q}\)	\(=\dfrac{150}{300-Q}\)
	\(\displaystyle \int d t\)	\(=\displaystyle \int \frac{150}{300-Q} d Q\)
	\( t\)	\(=-150\, \log _e(300-Q)+c\)

\( t\)	\(=150\, \log _e 295-150\, \log _e(300-Q)\)
\( t\)	\(=150\, \log _e\left(\dfrac{295}{300-Q}\right)\)

\(-2 y\)	\(=-2 \sqrt{3} x\)
\(y\)	\(=\sqrt{3} x\)