Band 5

Calculus, MET2 2025 VCAA 1

Let $g: R \rightarrow R$ be defined by $g(x)=4 x^3-3 x^4$.

Find the coordinates of both stationary points of $g$. (2 marks)

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Sketch the graph of $y=g(x)$ on the axes below, labelling the stationary points and axial intercepts with their coordinates. (2 marks)

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Complete the following gradient table with appropriate values of $x$ and $g^{\prime}(x)$ to show that $g$ has a stationary point of inflection. (2 marks)

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\begin{array}{|c|c|c|c|}
\hline \rule{0pt}{2.5ex}x \rule[-1ex]{0pt}{0pt}& \quad \quad \quad \quad& \quad \quad \quad \quad & \quad \quad \quad \quad\\
\hline\rule{0pt}{2.5ex} \quad g^{\prime}(x) \quad \rule[-1ex]{0pt}{0pt}& & & \\
\hline
\end{array}

Find the average value of $g$ between $x=0$ and $x=2$. (2 marks)
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Let $h$ be the result after applying a sequence of transformations to $g$, such that $h$ has a stationary point of inflection at $(1,0)$ and a local maximum at $(-1,1)$.
Write down a possible sequence of three transformations to map from $g$ to $h$. (3 marks)

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Let $X \sim \operatorname{Bi}(4, p)$ be a binomial random variable.
Show that $\operatorname{Pr}(X \geq 3)=g(p)$ for all $p \in[0,1]$. (2 marks)

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a. $\text{SP’s at} \ (0,0) \ \text{and}\ (1,1)$.

\begin{array}{|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}x \rule[-1ex]{0pt}{0pt}& \quad -1 \quad & \quad \quad 0 \quad \quad & \quad \quad \frac{1}{2} \quad \quad\\
\hline
\rule{0pt}{2.5ex} \quad g^{\prime}(x) \quad \rule[-1ex]{0pt}{0pt}& 24 & 0 & \frac{3}{2} \\
\hline
\end{array}

d. $I=-\dfrac{8}{5}$

e. $\text{Possible sequences include:}$

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text{Sequence 1}\rule[-1ex]{0pt}{0pt}&\text{Sequence 2} \\
\hline
\rule{0pt}{2.5ex}\text {1. Reflect in the } y \text {-axis.}\rule[-1ex]{0pt}{0pt}&\text{1. Dilate by factor 2 from the $y$-axis.} \\
\hline
\rule{0pt}{2.5ex}\text{2. Dilate by factor 2 from the $y$-axis.} \quad \rule[-1ex]{0pt}{0pt}& \text{2. Translate 1 unit left.} \\
\hline
\rule{0pt}{2.5ex}\text {3. Translate 1 unit right.}\quad \rule[-1ex]{0pt}{0pt}&\text{3. Reflect in the $y$-axis.} \\
\hline
\end{array}

\begin{array}{|l|l|}
\hline
\rule{0pt}{2.5ex}\text{Sequence 3}\rule[-1ex]{0pt}{0pt}&\text{Sequence 4} \\
\hline
\rule{0pt}{2.5ex}\text{1. Reflect in the $y$-axis.} \rule[-1ex]{0pt}{0pt}&\text{1. Translate 0.5 units left.} \\
\hline
\rule{0pt}{2.5ex}\text{2. Translate 0.5 units right.}\rule[-1ex]{0pt}{0pt}&\text{2. Dilate by factor 2 from the $y$-axis.} \\
\hline
\rule{0pt}{2.5ex}\text{3. Dilate by factor 2 from the $y$-axis.} \quad \rule[-1ex]{0pt}{0pt}&\text{3. Reflect in the $y$-axis.} \\
\hline
\end{array}

f. $X \sim \operatorname{Bi}(4, p)$

$\operatorname{Pr}(X \geqslant 3)$	$=\operatorname{Pr}(X=3)+\operatorname{Pr}(X=4)$
	$=4 p^3(1-p)+p^4$
	$=4 p^3-4 p^4+p^4$
	$=4 p^3-3 p^4$
	$=g(p)$

Show Worked Solution

a. $g(x)=4 x^3-3 x^4$

$g^{\prime}(x)=12 x^2-12 x^3=12 x^2(1-x)$

$\text{Solve} \ \ g^{\prime}(x)=0:$

$x=0,1$

$\text{SP’s at} \ (0,0) \ \text{and}\ (1,1)$.

b. $\text{Find} \ x \text {-intercepts:}$

$\text{Solve} \ \ g(x)=0 \ \ \Rightarrow\ \ x=0, \dfrac{4}{3}$

d. $\text{Solve integral (by CAS):}$

$I=\dfrac{1}{2-0} \displaystyle \int_0^2 g(x)\ d x=-\dfrac{8}{5}$

e. $\text{Possible sequences include:}$

f. $X \sim \operatorname{Bi}(4, p)$

$\operatorname{Pr}(X \geqslant 3)$	$=\operatorname{Pr}(X=3)+\operatorname{Pr}(X=4)$
	$=4 p^3(1-p)+p^4$
	$=4 p^3-4 p^4+p^4$
	$=4 p^3-3 p^4$
	$=g(p)$

♦ Mean mark (e) 40%.
Mean mark (f) 51%

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)

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Determine the $x$-coordinate of point $A$, in terms of $c$, when the distance $AB$ is a minimum. (2 marks)

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

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

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

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

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

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

Calculus, 2ADV C4 2025 MET2 17 MC

Given that $f: R \rightarrow R$ satisfies $\displaystyle \int_1^2 f(x)\, d x>\int_1^3 f(x)\, d x$, the graph of $y=f(x)$ could be

Show Answers Only

$A$

Show Worked Solution

$\displaystyle \int_1^2 f(x)\, d x>\int_1^3 f(x)\, d x$

$\text{To be true,} \ \displaystyle \int_2^3 f(x)\, d x<0 \ \ \text{(“net” area below} \ x \text{-axis)}$

$\Rightarrow A$

♦ Mean mark 38%.

Statistics, 2ADV 2025 MET2 14*

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.

Determine the exact value of $k$, expressing your answer in the form $a+b\sqrt{2}$, where $a, b \in R.$ (3 marks)

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

Show Worked Solution

$\displaystyle k \int_0^{\tfrac{\pi}{4}} \sin x+k \int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} \cos x$	$=1$
$k[-\cos x]_0^{\tfrac{\pi}{4}}+k[\sin x]_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}$	$=1$
$k\left(-\dfrac{1}{\sqrt{2}}+1\right)+k\left(1-\dfrac{1}{\sqrt{2}}\right)$	$=1$

$k\left(2-\dfrac{2}{\sqrt{2}}\right)=1$

$k\left(\dfrac{2 \sqrt{2}-2}{\sqrt{2}}\right)=1$

$k$	$=\dfrac{\sqrt{2}}{2 \sqrt{2}-2} \times \dfrac{2 \sqrt{2}+2}{2 \sqrt{2}+2}$
	$=\dfrac{4+2 \sqrt{2}}{8-4}$
	$=1+\dfrac{1}{2} \sqrt{2}$

Functions, 2ADV 2025 MET2 13 MC

The graphs of $y=f(x)$ and $y=g(x)$ are sketched on the same set of axes below.

Which of the following could be the graph of $y=g(f(x))$ ?

Show Answers Only

$C$

Show Worked Solution

$\text{Consider possible functions for each graph.}$

$y=f(x) \ \Rightarrow \ y=-x$

$y=g(x) \ \Rightarrow \ y=x(x+1)(x-1)(x-2)$

$y=g(f(x))$	$=-x(-x+1)(-x-1)(-x-2) $
	$=x(1-x)(x+1)(-x-2) $
	$=-x(1-x)(x+1)(x+2) $

$\text{By elimination:}$

$\text{Graph cuts \(x$-axis at}\ \ x=0, 1, -1, -2\ \ \text{(eliminate A and D)}\)

$\text{At}\ \ x=\dfrac{1}{2},\ g(f(x))<0\ \ \text{(eliminate B)}$

$\Rightarrow C$

♦ Mean mark 45%.

Functions, 2ADV 2025 MET2 10*

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

Determine the values of $\sin(x)$ where the inequality $f(g(x))>0$ is satisfied. (3 marks)

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$\dfrac{1}{2} \lt \sin(x) \leq 1$

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 \(f(g(x))=0$ for $a$:}\)

$a=\dfrac{-1 \pm \sqrt{1-4(2)(-1)}}{4}=-1, \ \dfrac{1}{2} $

$\text{Since}\ \ 2 a^2+a-1\ \ \text{is a concave up parabola:}$

$2 a^2+a-1>0\ \ \text{when}\ \ a<-1\ \cup\ a>\dfrac{1}{2} $

$\text{Since}\ \ -1\leq \sin(x) \leq 1 :$

$f(g(x))>0\ \ \text{when}\ \ \dfrac{1}{2} \lt \sin(x) \leq 1$

Probability, 2ADV S1 2025 MET2 9*

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, determine in terms of $m$ and $n$, the probability that they walked to school, giving your answer in its simplest form. (3 marks)

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

Show Worked Solution

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

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

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

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

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

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

Calculus, 2ADV C4 2025 MET2 6 MC

The trapezium rule is used, with two trapeziums, to estimate the area bounded by the graph of $y=f(x)$, the $x$-axis and the lines $x=0$ and $x=1$.

For which function will the trapezium rule estimate be larger than the exact area?

$f(x)=3-e^x$
$f(x)=x^3+1$
$f(x)=3 \sin (x)+1$
$f(x)=\log _e(x+3)$

Show Answers Only

$B$

Show Worked Solution

$\text{Rough sketch the shape of each graph.}$

$\text{Consider option B:}$

♦ Mean mark 50%.

$\text{By drawing trapeziums (see graph), the estimated area}$

$\text{is greater than the actual area.}$

$\Rightarrow B$

Functions, MET2 2025 VCAA 20 MC

Let $a>1$, and consider the functions $f$ and $g$ defined below.

\begin{align*}
\begin{aligned}
& f: R \rightarrow R, \ f(x)=a^x \\
& g: R \rightarrow R, \ g(x)=a^{2 x+2}
\end{aligned}
\end{align*}

Which one of the following sequences of transformations, when applied to $f(x)$, does not produce $g(x)$?

dilation by a factor of $\dfrac{1}{2}$ from the $y$-axis, then
translation by 1 unit in the negative direction of the $x$-axis
dilation by a factor of $\dfrac{1}{2}$ from the $y$-axis, then
dilation by a factor of $a^2$ from the $x$-axis
dilation by a factor of $a$ from the $x$-axis, then
dilation by a factor of $\dfrac{1}{2}$ from the $y$-axis, then
translation by 1 unit in the positive direction of the $x$-axis
dilation by a factor of $a^3$ from the $x$-axis, then
translation by 1 unit in the positive direction of the $x$-axis, then
dilation by a factor of $\dfrac{1}{2}$ from the $y$-axis

Show Answers Only

$C$

Show Worked Solution

$f(x)=a^x, \ g(x)=a^{2 x+2}$

$\text{Consider option C:}$

$\text{Dilate \(f(x)$ by a factor of $a$ from $x$-axis:}\)

$f_1(x)=a \times a^x=a^{x+1}$

$\text{Dilate \(f_1(x)$ by a factor of $\dfrac{1}{2}$ from $y$-axis:}\)

$f_2(x)=a^{2 x+1}$

$\text{Translate \(f_2(x)$ by 1 unit in positive direction of $x$-axis:}\)

$f_3(x)=a^{2\left(x+\frac{1}{2}-1\right)}=a^{2\left(x-\frac{1}{2}\right)}=a^{2 x-1} \neq a^{2 x+2}$

$\Rightarrow C$

♦♦ Mean mark 36%.

Statistics, MET2 2025 VCAA 18 MC

Consider the following graphs, which represent probability mass functions.

Which pair of these probability mass functions has the same mean?

$\text{I}$ and $\text{II}$
$\text{I}$ and $\text{IV}$
$\text{II}$ and $\text{III}$
$\text{II}$ and $\text{IV}$

Show Answers Only

$D$

Show Worked Solution

$E\left(X_1\right)=1 \times 0.1+2 \times 0.4+3 \times 0.4+4 \times 0.1=2.5$

$E\left(X_2\right)=1 \times 0.1+2 \times 0.2+3 \times 0.3+4 \times 0.4=3$

$E\left(X_3\right)=1 \times 0.45+2 \times 0.25+3 \times 0.15+4 \times 0.15=2$

$E\left(X_4\right)=1 \times 0.2+2 \times 0.2+3 \times 0.2+4 \times 0.2+5 \times 0.2=3$

$\text{Functions II and IV hare the same mean.}$

$\Rightarrow D$

♦ Mean mark 50%.

Calculus, MET2 2025 VCAA 17 MC

Given that $f: R \rightarrow R$ satisfies $\displaystyle \int_1^2 f(x)\, d x>\int_1^3 f(x)\, d x$, the graph of $y=f(x)$ could be

Show Answers Only

$A$

Show Worked Solution

$\displaystyle \int_1^2 f(x)\, d x>\int_1^3 f(x)\, d x$

$\text{To be true,} \ \displaystyle \int_2^3 f(x)\, d x<0 \ \ \text{(area below} \ x \text{-axis)}$

$\Rightarrow A$

♦ Mean mark 38%.

Functions, MET2 2025 VCAA 15 MC

The graph of $y=g(x)$ passes through the point $(1,3)$.

The graph of $y=1-g(2 x-3)$ must pass through the point

$(-1,-2)$
$(2,-2)$
$(-1,2)$
$(2,2)$

Show Answers Only

$B$

Show Worked Solution

$y=1-g(2 x-3)=-g\left(2\left(x-\frac{3}{2}\right)\right)+1$

$\text{Reflect \((1,3)$ in $x$-axis:}\)

$(1,3) \ \Rightarrow\ (1,-3)$

$\text{Dilate by a factor of} \ \frac{1}{2} \ \text{from} \ y \text {-axis: }$

$(1,-3) \Rightarrow\left(\dfrac{1}{2},-3\right)$

$\text{Translate 1 unit up and \(\frac{3}{2}$ units to right:}\)

$\left(\dfrac{1}{2},-3\right) \Rightarrow(2,-2)$

$\Rightarrow B$

♦ Mean mark 44%.

Functions, MET2 2025 VCAA 13 MC

The graphs of $y=f(x)$ and $y=g(x)$ are sketched on the same set of axes below.

Which of the following could be the graph of $y=(g \circ f)(x)$ ?

Show Answers Only

$C$

Show Worked Solution

$\text{Consider possible functions for each graph.}$

$y=f(x) \ \Rightarrow \ y=-x$

$y=g(x) \ \Rightarrow \ y=x(x+1)(x-1)(x-2)$

$\text{By CAS, sketch \(g(f(x))$ using the possible functions given above.}\)

$\Rightarrow C$

♦ Mean mark 45%.

ENGINEERING, PPT 2025 HSC 13 MC

Which row of the table correctly identifies the materials represented by the microstructures
shown in the top row of the table?

Show Answers Only

$D$

Show Worked Solution

Pearlite + $\alpha$ (ferrite) — large ferrite grains with pearlite patches = Mild steel; low carbon content means excess ferrite.
Pearlite + graphite flakes — distinctive dark flakes in a pearlite matrix = Grey cast iron; graphite precipitates as flakes during slow cooling.
Pearlite only — entirely pearlitic structure = High carbon steel; carbon content (~0.8%) produces near-100% pearlite (eutectoid).
Fe$_3$C + Pearlite — cementite network at grain boundaries = White cast iron; high carbon solidifies as iron carbide rather than graphite.

$\Rightarrow D$

♦ Mean mark 47%.

ENGINEERING, AE 2025 HSC 9 MC

The diagram shows an aircraft in ascent.

Which letter indicates the angle of attack of the aircraft?

Show Answers Only

$D$

Show Worked Solution

The angle of attack is the angle between the chord line of the wing and the oncoming airflow (relative airflow).
In the diagram, d is the angle between the chord line/flight path and the relative airflow direction — this is the angle of attack.
Angle a is the angle of ascent — measured between the flight path and the horizon.
Angles b and c relate to the flight path and vertical — neither represent the angle of attack.

$\Rightarrow D$

♦ Mean mark 35%.

ENGINEERING, TE 2025 HSC 4 MC

The screens of black and white cathode ray tube televisions were coated with a material that produced light when struck by electrons.

What was the material?

Plasma
Phosphor
Lead oxide
Magnesium oxide

Show Answers Only

$B$

Show Worked Solution

In a cathode ray tube (CRT), an electron gun fires a beam of electrons at high speed toward the screen.
The inner surface of the screen is coated with phosphor, a material that exhibits photoluminescence — it absorbs energy from the electron beam and re-emits it as visible light.
Plasma is a display technology distinct from CRT; lead oxide was used in the glass of the CRT envelope to absorb X-rays, not to produce light; magnesium oxide is not used in CRT screens.

$\Rightarrow B$

♦ Mean mark 45%.

Calculus, EXT1 C3 EQ-Bank 31

Given that $y(x)$ is a solution to the differential equation $\dfrac{d y}{d x}=x^2 y^3$, where $y(1)=3$, determine the domain of $y$. (4 marks)

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Show Answers Only

$x< \sqrt[3]{\dfrac{7}{6}}$

Show Worked Solution

$\dfrac{d y}{d x}=x^2 y^3$

$\displaystyle \int \frac{1}{y^3}\ d y$	$=\displaystyle \int x^2\ d x$
$-\dfrac{1}{2 y^2}$	$=\dfrac{1}{3} x^3+c$

$\text{Since} \ \ y(1)=3:$

$-\dfrac{1}{2(9)}=\dfrac{1}{3}+c \ \ \Rightarrow \ \ c=-\dfrac{1}{18}-\dfrac{1}{3}=-\dfrac{7}{18}$

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

$\text{Find domain of}\ y :$

$7-6 x^3$	$>0$
$6 x^3$	$<7$
$x^3$	$<\dfrac{7}{6}$
$x$	$< \sqrt[3]{\dfrac{7}{6}}$

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)

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

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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)
  
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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%.

Calculus, MET2 2025 VCAA 6 MC

The trapezium rule is used, with two trapeziums, to estimate the area bounded by the graph of $y=f(x)$, the $x$-axis and the lines $x=0$ and $x=1$.

For which function will the trapezium rule estimate be larger than the exact area?

$f(x)=3-e^x$
$f(x)=x^3+1$
$f(x)=3 \sin (x)+1$
$f(x)=\log _e(x+3)$

Show Answers Only

$B$

Show Worked Solution

$\text{Consider option B:}$

♦ Mean mark 50%.

$\text{By drawing trapeziums (see graph), the estimated area}$

$\text{is greater than the actual area.}$

$\Rightarrow B$

Algebra, MET2 2025 VCAA 4 MC

Consider the system of equations below containing the parameter $k$, where $k \in R$.

\begin{align*}
\begin{aligned}
k x+3 y & =k^2 \\
2 x+(2 k+1) y & =6-2 k
\end{aligned}
\end{align*}

Find the value(s) of $k$ for which this system has no real solutions.

$k=-2$ only
$k=\dfrac{3}{2}$ only
$k=-2$ or $\dfrac{3}{2}$
$k \in R \backslash\left\{-2, \dfrac{3}{2}\right\}$

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$A$

Show Worked Solution

$k x+3 y=k^2$

$2 x+(2 k+1) y=6-2 k$

♦ Mean mark 55%.

$\text{Find} \ k \ \text{where system has no real solutions.}$

$\dfrac{k}{2}=\dfrac{3}{2 k+1} \ \Rightarrow \ k=\dfrac{3}{2} \ \text { or } \ k=-2$

$\text{If} \ \ k=\dfrac{3}{2}:$

$\dfrac{k^2}{6-2 k}=\dfrac{3}{4}=\dfrac{k}{2} \ \ \text{(infinite solutions)}$

$\text{If} \ \ k=-2:$

$\dfrac{k^2}{6-2 k}=\dfrac{2}{5}, \dfrac{k}{2}=-1 \ \ \text {(no real solutions)}$

$\Rightarrow A$

Graphs, MET2 2025 VCAA 2 MC

All asymptotes of the graph of $y=2\, \tan \left(\pi\left(x+\dfrac{1}{2}\right)\right)$ are given by

$x=k, k \in Z$
$x=2 k, k \in Z$
$x=2 k+1, k \in Z$
$x=\dfrac{4 k+1}{2}, k \in Z$

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$A$

Show Worked Solution

$y=2\, \tan \left(\pi\left(x+\dfrac{1}{2}\right)\right)$

$\text{Since}\ 2\, \tan (x)\ \text {has asymptote at} \ \ x=\dfrac{\pi}{2}:$

$\text{Find \(x$ when}\)

$\pi\left(x+\dfrac{1}{2}\right)=\dfrac{\pi}{2} \ \ \Rightarrow \ \ x=0$

$\text{Period}=1$

$\therefore \ \text{General solution:} \ x=k, k \in Z$

$\Rightarrow A$

♦ Mean mark 51%.

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)

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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)

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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)

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

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What value does the quantity of salt in the tank approach as time approaches infinity?
Give your answer in kilograms. (1 mark)

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Find the time taken for the quantity of salt in the tank to reach 100 kg. (1 mark)

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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)

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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%.

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)

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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)
  
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2. Find the volume of the solid of revolution correct to two decimal places. (1 mark)
  
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Find the equations of the vertical asymptotes of the curve given by $y=\dfrac{3 x}{x^3-5 x+2}$. (1 mark)

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

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

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

Vectors, EXT2 V1 2025 SPEC2 18 MC

The lines given by ${\underset{\sim}{r}}_1(\lambda)=2 \underset{\sim}{i}+r \underset{\sim}{j}-3 \underset{\sim}{k}+\lambda(\underset{\sim}{i}-\underset{\sim}{j}+4 \underset{\sim}{k})$ and ${\underset{\sim}{r}}_2(\mu)=\underset{\sim}{i}+s \underset{\sim}{k}+\mu(\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k})$ intersect at the point $(4,3, t)$, where $\lambda, \mu \in R$ and $r, s$ and $t$ are real constants.

The values of $r, s$ and $t$ respectively are

2, 3 and 5
5, 3 and 5
5, 5 and 8
5, 8 and 5

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$D$

Show Worked Solution

\({\underset{\sim}{r}}_1(\lambda)=2\underset{\sim}{i}+r\underset{\sim}{j}-3\underset{\sim}{k}+\lambda\left(\underset{\sim}{i}-\underset{\sim}{j}+4\underset{\sim}{k}\right)\)

${\underset{\sim}{r}}_2(\mu)=\underset{\sim}{i}+0 \underset{\sim}{j}+s \underset{\sim}{k}+\mu(\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k})$

♦ Mean mark 49%.

$\text{Since intersection occurs at}\ (4,3, t):$

$4=2+\lambda \ \Rightarrow \ \lambda=2$

$3=r-\lambda \ \Rightarrow \ 3=r-2\ \Rightarrow\ r=5$

$t=-3+4 \lambda=-3+4(2)=5$

$4=1+\mu \ \Rightarrow \ \mu=3$

$t=s-\mu \ \Rightarrow \ 5=s-3\ \Rightarrow\ s=8$

$\Rightarrow D$

Statistics, SPEC2 2025 VCAA 20 MC

Let $P \sim N\left(-2,2^2\right), Q \sim N\left(3,3^2\right), R \sim N\left(5,6^2\right)$ and $Z \sim N(0,1)$.

Given that $P, Q$ and $R$ are independent random variables, $\operatorname{Pr}(3 P+2 Q-R>25)$ is equal to

$\operatorname{Pr}\left(Z>\dfrac{5 \sqrt{3}}{3}\right)$
$\operatorname{Pr}(Z>5)$
$\operatorname{Pr}\left(Z>\dfrac{5 \sqrt{66}}{11}\right)$
$\operatorname{Pr}\left(Z>\dfrac{30}{7}\right)$

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$A$

Show Worked Solution

$P \sim N\left(-2,2^2\right), Q \sim N\left(3,3^2\right), R \sim N\left(5,6^2\right)$

$\text{Find} \ \ \operatorname{Pr}(3 P+2Q-R>25):$

$E(3 P+2 Q-R)=3 E(P)+2 E(Q)-E(R)=-5$

$\operatorname{Var}(3 P+2 Q-R)=9 \operatorname{Var}(P)+4 \operatorname{Var}(Q)+\operatorname{Var}(R)=108$

$\operatorname{Pr}(3 \operatorname{P}+2 Q-R>25)=\operatorname{Pr}\left(Z>\dfrac{25-(-5)}{\sqrt{108}}\right)=\operatorname{Pr}\left(Z>\dfrac{5 \sqrt{3}}{3}\right)$

$\Rightarrow A$

Vectors, SPEC2 2025 VCAA 18 MC

The values of $r, s$ and $t$ respectively are

2, 3 and 5
5, 3 and 5
5, 5 and 8
5, 8 and 5

Show Answers Only

$D$

Show Worked Solution

\({\underset{\sim}{r}}_1(\lambda)=2\underset{\sim}{i}+r\underset{\sim}{j}-3\underset{\sim}{k}+\lambda\left(\underset{\sim}{i}-\underset{\sim}{j}+4\underset{\sim}{k}\right)\)

${\underset{\sim}{r}}_2(\mu)=\underset{\sim}{i}+0 \underset{\sim}{j}+s \underset{\sim}{k}+\mu(\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k})$

♦ Mean mark 49%.

$\text{Since intersection occurs at}\ (4,3, t):$

$4=2+\lambda \ \Rightarrow \ \lambda=2$

$3=r-\lambda \ \Rightarrow \ 3=r-2\ \Rightarrow\ r=5$

$t=-3+4 \lambda=-3+4(2)=5$

$4=1+\mu \ \Rightarrow \ \mu=3$

$t=s-\mu \ \Rightarrow \ 5=s-3\ \Rightarrow\ s=8$

$\Rightarrow D$

Vectors, SPEC2 2025 VCAA 17 MC

The acceleration vector of a particle that starts from rest is given by

$\underset{\sim}{ a }(t)=4 \cos (2 t) \underset{\sim}{ i }+10 \sin (2 t) \underset{\sim}{ j }-6 e^{-2 t} \underset{\sim}{ k }$, where $t \geq 0$.

The velocity vector of the particle, $v(t)$, is given by

$\underset{\sim}{ v }(t)=2 \sin (2 t) \underset{\sim}{ i }-5 \cos (2 t) \underset{\sim}{ j }+3 e^{-2 t} \underset{\sim}{ k }$
$\underset{\sim}{ v }(t)=2 \sin (2 t) \underset{\sim}{ i }-5(\cos (2 t)+1) \underset{\sim}{ j }+3\left(e^{-2 t}+1\right) \underset{\sim}{ k }$
$\underset{\sim}{ v }(t)=2 \sin (2 t) \underset{\sim}{ i }-5(\cos (2 t)-1) \underset{\sim}{ j }+3\left(e^{-2 t}-1\right) \underset{\sim}{ k }$
$\underset{\sim}{ v }(t)=-8 \sin (2 t) \underset{\sim}{ i }+20 \cos (2 t) \underset{\sim}{ j }+12 e^{-2 t} \underset{\sim}{ k }$

Show Answers Only

$C$

Show Worked Solution

$\underset{\sim}{a}(t)=4 \cos (2 t) \underset{\sim}{i}+10 \sin (2 t) \underset{\sim}{j}-6 e^{-2 t} \underset{\sim}{k}$

$\underset{\sim}{v}(t)=\displaystyle \int \underset{\sim}{a}(t)=2 \sin (2 t) \underset{\sim}{i}-5 \cos (2 t) \underset{\sim}{j}+3 e^{-2 t} \underset{\sim}{k}+\underset{\sim}{c}$

♦ Mean mark 52%.

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

$0=0-5 \underset{\sim}{j}+3 \underset{\sim}{k}+\underset{\sim}{c} \ \Rightarrow \ \underset{\sim}{c}=5 \underset{\sim}{j}-3 \underset{\sim}{k}$

$v(t)$	$=2 \sin (2 t) \underset{\sim}{i}-5 \cos (2 t) \underset{\sim}{j}+3 e^{-2 t} \underset{\sim}{k}+5\underset{\sim}{j}-3 \underset{\sim}{k}$
	$=2 \sin (2 t) \underset{\sim}{i}-5(\cos (2 t)-1) \underset{\sim}{j}+3\left(e^{-2 t}-1\right) \underset{\sim}{k}$

$\Rightarrow C$

Vectors, SPEC2 2025 VCAA 15 MC

Consider the two planes described by the equations $2 x+2 y+z=2$ and $a x+4 z=1$, where $a$ is a positive constant.

The angle between the two planes is $\cos ^{-1}\left(\dfrac{2}{3}\right)$.

The value of $a$ satisfies the equation

$a+2=\sqrt{a^2+16}$
$\dfrac{2 a+4}{3 \sqrt{a^2+16}}=\dfrac{3}{2}$
$2 a+4=3 \sqrt{a^2+16}$
$\dfrac{2 a+4}{\sqrt{a^2+16}}=\dfrac{2}{3}$

Show Answers Only

$A$

Show Worked Solution

$\text{Plane 1:}\ \ 2 x+2 y+z=2$

$\text{Plane 2:}\ \ a x+4 z=1$

$\text{Angle between planes = angle between normals}$

♦ Mean mark 52%.

$\displaystyle {\underset{\sim}{n}}_1=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right), \quad {\underset{\sim}{n}}_2=\left(\begin{array}{l}a \\ 0 \\ 4\end{array}\right)$

${\underset{\sim}{n}}_1 \cdot {\underset{\sim}{n}}_2$	$=\abs{{\underset{\sim}{n}}_1}\abs{{\underset{\sim}{n}}_2}\cos(\theta)$
$2 a+4$	$=\sqrt{4+4+1} \times \sqrt{a^2+16} \times \dfrac{2}{3}$
$2 a+4$	$=3 \sqrt{a^2+16} \times \dfrac{2}{3}$
$a+2$	$=\sqrt{a^2+16}$

$\Rightarrow A$

Vectors, SPEC2 2025 VCAA 14 MC

For non-zero vectors $\underset{\sim}{a}$ and $\underset{\sim}{b}$, if $\underset{\sim}{a} \cdot \underset{\sim}{b}=\abs{\underset{\sim}{a} \times \underset{\sim}{b}}$, then the angle between $\underset{\sim}{a}$ and $\underset{\sim}{b}$ is

$0$
$\dfrac{\pi}{4}$
$\dfrac{\pi}{2}$
$\dfrac{3 \pi}{4}$

Show Answers Only

$B$

Show Worked Solution

$\text{Using the vector definitions:}$

$\underset{\sim}{a} \cdot \underset{\sim}{b}=\abs{\underset{\sim}{a}}\abs{\underset{\sim}{b}}\cos (\theta)\ \ldots\ (1)$

$\underset{\sim}{a} \times \underset{\sim}{b}=\abs{\underset{\sim}{a}}\abs{\underset{\sim}{b}}\sin (\theta)\ \ldots\ (2)$

$\text{Given}\ (1)=(2):$

$\cos (\theta)=|\sin (\theta)|$

$\tan (\theta)=1$

$\Rightarrow B$

♦ Mean mark 48%.

Calculus, SPEC2 2025 VCAA 13 MC

From an open window, a person projects a ball vertically up using an outstretched arm so the ball does not strike any part of the building. The point of projection of the ball is 50 m above the ground and its velocity of projection is 20 m s$^{-1}$.

The time, in seconds, it takes for the ball to reach the tray of a truck that is 1 m above the ground directly below the point of projection is closest to

1.72
5.80
5.83
1.75

Show Answers Only

$B$

Show Worked Solution

$u=20\ \text{ms}^{-1}, \ a=-9.8\ \text{ms}^{-2}, \ s=-49$

$\text{Find}\ t\ \text{when}\ \ s=-49:$

$\text{Using}\ \ s=u t+\dfrac{1}{2} a t^2,$

$-49=20 t-4.9 t^2 \ \ \Rightarrow \ \ t=-1.72,5.80$

$\Rightarrow B$

Mean mark 55%.

Calculus, SPEC2 2025 VCAA 12 MC

A particle moves along a straight line with constant acceleration. It passes through a point $A$ with velocity $u$ m s$^{-1}$ and then through a point $B$ with velocity $v$ ms$^{-1}$.

The velocity of the particle at the midpoint of the line segment $AB$ is given by

$\dfrac{u+v}{2}$
$u+\dfrac{u+v}{2}$
$\dfrac{u^2+v^2}{2}$
$\sqrt{\dfrac{u^2+v^2}{2}}$

Show Answers Only

$D$

Show Worked Solution

$\text{Let} \ \ u=\text{velocity at}\ A, \ v=\text {velocity at} \ B$

$v^2=u^2+2 a s \ \Rightarrow \ 2 a s=v^2-u^2\ \ldots\ (1)$

$\text{At halfway point:}$

$v_m^2=u^2+2 a \times \dfrac{s}{2}$

$\text {Using (1) above:}$

$v_m^2=u^2+\frac{1}{2}\left(v^2-u^2\right)=\dfrac{u^2+v^2}{2}$

$\Rightarrow D$

♦♦ Mean mark 33%.

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)

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Find the initial acceleration, in terms of $k$, of the particle in m s$^{-2}$. (2 marks)

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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)

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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%.

Calculus, SPEC2 2025 VCAA 9 MC

A parametric curve is given by $x=k t, y=e^{k t}$, where $k$ is a positive constant. The curve is rotated about the $x$-axis from $t=a$ to $t=b$, where $b>a$, to form a surface of revolution.

The area of this surface is given by

$2 \displaystyle \pi \int_a^b e^{k t} \sqrt{k^2 t^2+e^{2 k t}}\ d t$
$2 \displaystyle\pi \sqrt{k} \int_a^b e^{k t} \sqrt{1+e^{k t}}\ d t$
$2 \displaystyle\pi \int_{\Large{e^{k a}}}^{\Large{e^{k b}}} \sqrt{1+u^2}\ d u$
$2 \displaystyle\pi \int_{\Large{e^{k a}}}^{\Large{e^{k b}}} u \sqrt{1+u^2}\ d u$

Show Answers Only

$C$

Show Worked Solution

$\text{Using formula for S.A. of a curve rotated about \(x$-axis:}\)

$\text{S.A. }$	$=2 \pi \displaystyle \int_a^b x \sqrt{\left(\frac{d x}{d t}\right)^2+\left(\frac{d y}{d t}\right)^2} \, dt$
	$=2 \pi \displaystyle\int_a^b e^{k t} \sqrt{k^2+k^2 e^{2 k t}} \, dt$

♦ Mean mark 49%.

$\text{Using substitution} \ \ u=e^{kt}:$

$\text{S.A.}=2 \pi \displaystyle\int_{e^{ka}}^{e^{k b}} \frac{1}{k} \sqrt{k^2\left(1+u^2\right)} \, du$

$\Rightarrow C$

Calculus, SPEC2 2025 VCAA 2 MC

Consider the following statement.

'If $f^{\prime \prime}(0)=0$, then the graph of $f$ necessarily has a point of inflection at $x=0$.'

A counter-example that disproves this statement is when

$f(x)=\sin ^{-1}(x)$
$f(x)=\dfrac{2 x}{x^2-1}$
$f(x)=x^{\tfrac{1}{3}}$
$f(x)=x^4-x$

Show Answers Only

$D$

Show Worked Solution

$\text{POI exists at} \ \ x=0 \ \ \text{if:}$

$f^{\prime \prime}(0)=0\ \text{and concavity changes either side of \(\ x=0$ ($\text{i.e.}\ f^{\prime \prime}$ changes sign)}\)

$\text{Inspect each graph (by CAS).}$

$\Rightarrow D$

♦ Mean mark 48%.

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)

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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)

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Sketch the graph of $y=f(x)$ on the axes below.
Label the asymptotes with their equations. (3 marks)

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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%.

Vectors, SPEC1 2025 VCAA 5

The position vectors of particles $P$ and $Q$ at time $t$ seconds are given by

\begin{align*}
{\underset{\sim}{r}}_P(t)=\left(t^3+a t^2\right) \underset{\sim}{i}-\underset{\sim}{j} \ \ \text {and} \ \ {\underset{\sim}{r}}_Q(t)=(b t+2 t) \underset{\sim}{i}+\left(2 t^2+c t+t\right) \underset{\sim}{j},
\end{align*}

where $t \geq 0$ and $a, b, c \in R$.

The particles collide when $t=1$.

Show that for collision to occur when $t=1$, the value of $c$ is $-4$. (1 mark)

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When the particles collide, their velocities are at right angles to each other.

Find the two possible values of $a$ for collision to occur when $t=1$. (2 marks)

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When the particles collide at $t=1$, the magnitudes of their accelerations are equal.
Find the values of $a$ and $b$ for collision to occur when $t=1$. (1 mark)

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a. $\text{\(P$ and $Q$ collide at $\ t=1$.}\)

$\text{Equate co-efficients of} \ \underset{\sim}{j}:$

$2+c+1=-1 \ \ \Rightarrow \ \ c=-4$

b. $a=-\dfrac{3}{2},-1$

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

Show Worked Solution

a. $\text{\(P$ and $Q$ collide at $\ t=1$.}\)

$\text{Equate co-efficients of} \ \underset{\sim}{j}:$

$2+c+1=-1 \ \ \Rightarrow \ \ c=-4$

b. $r_P=\left(t^3+a t^2\right)\underset{\sim}{i}-\underset{\sim}{j} \ \ \Rightarrow \ \ \dot{r}_P=\left(3 t^2+2 a t\right) \underset{\sim}{i}$

$r_Q=(b t+2 t) \underset{\sim}{i}+\left(2 t^2-3 t\right)\underset{\sim}{j} \ \ \Rightarrow \ \ \dot{r}_Q=(b+2)\underset{\sim}{i}+(4 t-3)\underset{\sim}{j}$

$\text {Velocities are perpendicular at} \ \ t=1:$

$\displaystyle \dot{r}_p \cdot \dot{r}_Q=\binom{3+2 a}{0}\binom{b+2}{1}=(3+2 a)(b+2)=0\ \ldots\ (1)$

♦ Mean mark (b) 49%.

$\text{Equating co-efficients of \(\underset{\sim}{i}$ at $\ t=1$:}\)

$1+a=b+2$

$\text{Substitute into (1) above:}$

$(3+2 a)(1+a)=0$

$a=-\dfrac{3}{2},-1$

c. $\ddot{r}_P=(6 t+2 a)\underset{\sim}{i}, \ \ddot{r}_Q=4 \underset{\sim}{j}$

$\text{At \(\ t=1$, magnitudes of $\ddot{r}_P$ and $\ddot{r}_Q$ are equal.}\)

$6+2 a=4 \ \ \Rightarrow \ \ a=-1$

$1-1=b+2 \ \ \Rightarrow \ \ b=-2$

♦♦♦ Mean mark (c) 25%.

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)

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Find the initial acceleration, in terms of $k$, of the particle in m s$^{-2}$. (2 marks)

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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)

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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%.

Probability, 2ADV S3 2025 MET1 8

Consider

\begin{align*}
f(x)=\left\{\begin{array}{cc}
\dfrac{3}{8}(4-3 x) & 0 \leq x \leq \dfrac{4}{3} \\
0 & \text {otherwise }
\end{array}\right.
\end{align*}

The continuous random variable $X$ has probability density function $f(x)$.
Find $k$ such that $\operatorname{Pr}(X>k)=\dfrac{9}{16}$. (3 marks)

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The function $h(x)$ is a transformation of $f(x)$ such that
\begin{align*}
\ \ \ \ h(x)=m f(x)+n
\end{align*}
where $m$ and $n$ are real numbers.
Find $\displaystyle \int_0^{\tfrac{4}{3}} h(x) d x$ in terms of $m$ and $n$. (2 marks)

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Show Answers Only

a. $k=\dfrac{1}{3}$

b. $\displaystyle\int_0^{\frac{4}{3}} h(x)=m+\dfrac{4}{3} n$

Show Worked Solution

a. $\text{Solve}\ \ \operatorname{Pr}(X>k)=\dfrac{9}{16}\ \ \text{for}\ k$ :

$\operatorname{Pr}(X>k)$	$=\displaystyle\int_k^{\frac{4}{3}} \frac{3}{8}(4-3 x) d x$
	$=\displaystyle-\frac{3}{8} \int_k^{\frac{4}{3}}(3 x-4) d x$
	$=-\dfrac{3}{8} \cdot \dfrac{1}{2} \cdot \dfrac{1}{3}\left[(3 x-4)^2\right]_k^{\frac{4}{3}}$
	$=-\dfrac{1}{16}\left[(4-4)^2-(3 k-4)^2\right]$
	$=\dfrac{(3 k-4)^2}{16}$

♦ Mean mark (a) 47%.

$\dfrac{(3 k-4)^2}{16}$	$=\dfrac{9}{16}$
$(3 k-4)^2$	$=9$
$3 k-4$	$= \pm 3$

$3 k=1 \ \ \ \ \text{or}\ \ \ \ 3 k=7$

$k=\dfrac{1}{3} \quad \quad \ \ \ k=\dfrac{7}{3}\ \ \left(\text{No solution as}\ k \in\left[0, \dfrac{4}{3}\right]\right)$

$\therefore\ k=\dfrac{1}{3}$

b. $h(x)=m f(x)+n$

	$\displaystyle\int_0^{\frac{4}{3}} h(x)$	$=\displaystyle \int_0^{\frac{4}{3}}(m f(x)+n) d x$
		$=m \underbrace{\displaystyle \int_0^{\frac{4}{3}} f(x) d x}_{=1\ \ \text{since p.d.f. }}+n \displaystyle\int_0^{\frac{4}{3}} 1\ d x$
		$=m+n[x]_0^{\frac{4}{3}}$
		$=m+\dfrac{4}{3} n$

♦♦ Mean mark (b) 38%.

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)

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Solve $f(x)=0$ for $x$. (3 marks)

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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)

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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%.

Algebra, MET1 2025 VCAA 9

Consider the functions

$f: R \backslash\{1\} \rightarrow R, f(x)=\dfrac{w^2}{(x-1)^2}$

and

$g: R \rightarrow R, g(x)=(x-w)^2$

where $w \in R$.

If $w=-3$, find the four solutions to $f(x)=g(x)$. (3 marks)

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Consider the case where $w>0$.
1. Find, in terms of $w$, the coordinates of the minimum point of the graph of $y=(x-1)(x-w)$. (2 marks)
  
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2. Hence, or otherwise, find the positive values of $w$ for which $f(x)=g(x)$ has exactly three solutions. (2 marks)
  
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a. $\text{If} \ \ w=-3, \text{find 4 solutions to} \ \ f(x)=g(x):$

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

$(x+3)^2(x-1)^2=9$

$(x+3)(x-1)= \pm 3$

$\text{Case 1}$

$x^2+2 x-3=3$

$x^2+2 x-6=0$

$x=-1 \pm \sqrt{7}$

$\text{Case 2}$

$x^2+2 x-3$	$=-3$
$x(x+2)$	$=0$
$x=0 \ \ \text{or} \ -2$

b.i. $\text{Min TP at} \ \left(\dfrac{w+1}{2}, -\dfrac{1}{4} (w-1)^2\right)$

b.ii. $\text{See worked solutions.}$

Show Worked Solution

a. $\text{If} \ \ w=-3, \text{find 4 solutions to} \ \ f(x)=g(x):$

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

$(x+3)^2(x-1)^2=9$

$(x+3)(x-1)= \pm 3$

$\text{Case 1}$

$x^2+2 x-3=3$

$x^2+2 x-6=0$

$x=-1 \pm \sqrt{7}$

♦♦ Mean mark (a) 34%.

$\text{Case 2}$

$x^2+2 x-3$	$=-3$
$x(x+2)$	$=0$
$x=0 \ \ \text{or} \ -2$

b.i. $\text{Find minimum TP of} \ \ y=(x-1)(x-w)$

$\text{Axis of quadratic:}\ \ x=\dfrac{w+1}{2},\ \ (w>0)$

$y$	$=\left(\dfrac{w+1}{2}-1\right)\left(\dfrac{w+1}{2}-w\right)$
	$=\left(\dfrac{w-1}{2}\right)\left(\dfrac{1-w}{2}\right)$
	$=-\dfrac{1}{4}(w-1)^2$

$\therefore \ \text{Min TP at} \ \left(\dfrac{w+1}{2}, -\dfrac{1}{4} (w-1)^2\right)$

♦♦ Mean mark (b.i) 39%.
♦♦♦ Mean mark (b.ii) 11%.

b.ii. $\text{Solve for \(w\ (w>0)$ where $f(x)=g(x)$:}\)

$\dfrac{w^2}{(x-1)^2}$	$=(x-w)^2$
$(x-w)^2(x-1)^2$	$=w^2$
$[(x-w)(x-1)]^2-w^2$	$=0$

$\text{Difference between two squares:}$

$[(x-w)(x-1)-w][(x-w)(x-1)+w]=0$

$\text{Case 1}$

$(x-w)(x-1)-w$	$=0$
$x^2-x-w x+w-w$	$=0$
$x^2-(w+1) x$	$=0$
$x[x-(w+1)]$	$=0$

$x=0 \ \ \text{or} \ \ w+1$

$\text{Case 2}$

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

$\text{1 solution if} \ \ \Delta=0:$

$\Delta$	$=[-(w+1)]^2-4 \times 1 \times 2 w$
	$=w^2+2 w+1-8 w$
	$=w^2-6 x+1$

$\text{If} \ \ \Delta=0:$

$w$	$=\dfrac{6 \pm \sqrt{(-6)^2-4 \times 1 \times 1}}{2}$
	$=\dfrac{6 \pm \sqrt{32}}{2}$
	$=3+2 \sqrt{2}\ \ (\text{reject}\ w=3-2 \sqrt{2}\ \ \text{as}\ \ w>0)$

$\therefore f(x)=g(x) \ \ \text{has exactly 3 solutions}\ (w>0).$

Probability, MET1 2025 VCAA 8

Consider

\begin{align*}
f(x)=\left\{\begin{array}{cc}
\dfrac{3}{8}(4-3 x) & 0 \leq x \leq \dfrac{4}{3} \\
0 & \text {otherwise }
\end{array}\right.
\end{align*}

The continuous random variable $X$ has probability density function $f(x)$.
Find $k$ such that $\operatorname{Pr}(X>k)=\dfrac{9}{16}$. (3 marks)

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The function $h(x)$ is a transformation of $f(x)$ such that
\begin{align*}
\ \ \ \ h(x)=m f(x)+n
\end{align*}
where $m$ and $n$ are real numbers.
Find $\displaystyle \int_0^{\tfrac{4}{3}} h(x) d x$ in terms of $m$ and $n$. (2 marks)

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Show Answers Only

a. $k=\dfrac{1}{3}$

b. $\displaystyle\int_0^{\frac{4}{3}} h(x)=m+\dfrac{4}{3} n$

Show Worked Solution

a. $\text{Solve}\ \ \operatorname{Pr}(X>k)=\dfrac{9}{16}\ \ \text{for}\ k$ :

$\operatorname{Pr}(X>k)$	$=\displaystyle\int_k^{\frac{4}{3}} \frac{3}{8}(4-3 x) d x$
	$=\displaystyle-\frac{3}{8} \int_k^{\frac{4}{3}}(3 x-4) d x$
	$=-\dfrac{3}{8} \cdot \dfrac{1}{2} \cdot \dfrac{1}{3}\left[(3 x-4)^2\right]_k^{\frac{4}{3}}$
	$=-\dfrac{1}{16}\left[(4-4)^2-(3 k-4)^2\right]$
	$=\dfrac{(3 k-4)^2}{16}$

♦ Mean mark (a) 47%.

$\dfrac{(3 k-4)^2}{16}$	$=\dfrac{9}{16}$
$(3 k-4)^2$	$=9$
$3 k-4$	$= \pm 3$

$3 k=1 \ \ \ \ \text{or}\ \ \ \ 3 k=7$

$k=\dfrac{1}{3} \quad \quad \ \ \ k=\dfrac{7}{3}\ \ \left(\text{No solution as}\ k \in\left[0, \dfrac{4}{3}\right]\right)$

$\therefore\ k=\dfrac{1}{3}$

b. $h(x)=m f(x)+n$

	$\displaystyle\int_0^{\frac{4}{3}} h(x)$	$=\displaystyle \int_0^{\frac{4}{3}}(m f(x)+n) d x$
		$=m \underbrace{\displaystyle \int_0^{\frac{4}{3}} f(x) d x}_{=1\ \ \text{since p.d.f. }}+n \displaystyle\int_0^{\frac{4}{3}} 1\ d x$
		$=m+n[x]_0^{\frac{4}{3}}$
		$=m+\dfrac{4}{3} n$

♦♦ Mean mark (b) 38%.

Calculus, MET1 2025 VCAA 7

Let $f: R \rightarrow R, f(x)=x^3-x^2-16 x-20$.

Verify that $x=5$ is a solution of $f(x)=0$. (1 mark)

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Express $f(x)$ in the form $(x+d)^2(x-5)$, where $d \in R$. (2 marks)

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Consider the graph of $y=f(x)$, as shown below.
Complete the coordinate pairs of all axial intercepts of $y=f(x)$. (1 mark)

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Let $g: R \rightarrow R, g(x)=x+2$.
1. State the coordinates of the stationary point of inflection for the graph of $y=f(x) g(x)$. (1 mark)
  
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2. Write down the values of $x$ for which $f(x) g(x) \geq 0$. (1 mark)
  
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a. $f(5)=5^3-5^2-16 \times 5-20 =0$

b. $f(x)=(x+2)^2(x-5)$

d.i. $(-2,0)$

d.ii. $x \in(-\infty,-2] \cup[5, \infty)$

Show Worked Solution

a. $f(x)=x^3-x^2-16 x-20$

$f(5)=5^3-5^2-16 \times 5-20=125-25-80-20=0$

$\therefore x=5\ \ \text{is a solution of}\ f(x)$.

b. $\text{By long division:}$

$f(x)=\left(x^2+4 x+4\right)(x-5)=(x+2)^2(x-5)$

d.i.	$y$	$=f(x) \cdot g(x)$
		$=(x+2)^2(x-5)(x+2)$
		$=(x+2)^3(x-5)$

$(x+2)^3\ \text{factor}\ \Rightarrow\ \text{SP of inflection at} \ (-2,0)$

♦ Mean mark (d.i) 44%.
♦♦ Mean mark (d.ii) 27%.

d.ii.

$\text{By inspection of graph:}$

$f(x) g(x) \geqslant 0 \ \ \text{when}\ \ x \leqslant-2\ \cup\ x \geqslant 5$

$x \in(-\infty,-2] \cup[5, \infty) \ \text{also correct.}$

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)

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Outline how neural pathways allow the hypothalamus to maintain body temperature within normal range. (3 marks)

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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)

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Solve $f(x)=0$ for $x$. (3 marks)

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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)

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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%.

Networks, GEN2 2025 VCAA 18

Frances is constructing a home gym. This project requires 12 activities, $A$ to $L$, to be completed.

The activity network below shows each activity and its completion time in days.

This network contains two critical paths.
State the activities that are common to both critical paths. (1 mark)

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Determine the latest start time, in days, for activity $E$. (1 mark)

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Which activity has the longest float time? (1 mark)

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The table below shows five activities that can have their completion time reduced.
It shows the maximum reduction time (days) and the additional cost per day, for each of the five activities.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ \textbf{Activity} \ \ & \textbf{Maximum reduction time} & \textbf{Additional cost per day }\\
\textbf{} & \textbf{(days)} \rule[-1ex]{0pt}{0pt} & \textbf{(\$) }\\
\hline
\rule{0pt}{2.5ex} \textit{A} \rule[-1ex]{0pt}{0pt} & \text{2} \rule[-1ex]{0pt}{0pt} & \text{500} \\
\hline
\rule{0pt}{2.5ex} \textit{F} \rule[-1ex]{0pt}{0pt} & \text{4} \rule[-1ex]{0pt}{0pt} & \text{150} \\
\hline
\rule{0pt}{2.5ex} \textit{G} \rule[-1ex]{0pt}{0pt} & \text{4} \rule[-1ex]{0pt}{0pt} & \text{150} \\
\hline
\rule{0pt}{2.5ex} \textit{H} \rule[-1ex]{0pt}{0pt} & \text{2} \rule[-1ex]{0pt}{0pt} & \text{300} \\
\hline
\rule{0pt}{2.5ex} \textit{K} \rule[-1ex]{0pt}{0pt} & \text{1} \rule[-1ex]{0pt}{0pt} & \text{100} \\
\hline
\end{array}

Frances would like to construct the home gym in three days less than was previously possible.
What is the minimum additional amount Frances will need to pay? (1 mark)

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a. $\text{Critical paths: }ADHJL, ADIKL$

b. $\text{LST for Activity}\ E=9 \ \text{days}$

c. $\text{Activity \(F$.}\)

d. $\$1400$

Show Worked Solution

a. $\text{Scan network forwards/backwards:}$

$\text{Critical paths: }ADHJL, ADIKL$

b. $\text{LST for Activity}\ E$

$=20-4-4-3$

$=9 \ \text{days}$

♦ Mean mark (a) 44%.
♦ Mean mark (b) 46%.

c.	$\begin{array}{\|l\|c\|c\|c\|c\|c\|} \hline \rule{0pt}{2.5ex}\ \ \text{Activity} \ \ \rule[-1ex]{0pt}{0pt}& \ \ B \ \ &\ \ C \ \ & \ \ E \ \ & \ \ F \ \ & \ \ G \ \ \\ \hline \rule{0pt}{2.5ex} \ \ \text{Float} \ \ \rule[-1ex]{0pt}{0pt}& 5 & 1 & 5 & 6 & 1 \\ \hline \end{array}$

$\therefore\ \text{Activity \(F$ has the longest float time.}\)

♦ Mean mark (c) 47%.

d. $\text{Current time = 25 days}$

$\text{New time = 22 days.}$

$\text{Reduce:} \ A(2 \ \text {days}), H(1 \ \text{day}), K(1 \ \text{day})$

$\text{Minimum cost}=2 \times 500+1 \times 300+1 \times 100=\$1400$

♦♦♦ Mean mark (d) 21%.

Networks, GEN2 2025 VCAA 17

A gym is hosting a competition in which competitors will complete activities at eight different stations.

On the network below, the vertices represent the stations. The edges represent the walkways between the stations and the numbers show the distance, in metres, between them.

The gym owner would like to make sure that all the walkways are clear before the competition starts.

The gym owner would like to begin and end the inspection of the walkways at station $A$.

Explain, with reference to the degrees of the vertices, why the gym owner's intended route must involve some repeated edges. (1 mark)

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What is the minimum distance, in metres, that the gym owner will cover when completing the inspection? (1 mark)

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a. $\text{Vertices \(C$ and $E$ are of odd degree.}\)

$\text{The circuit described is Eulerian which requires all vertices to be}$

$\text{of even degree if edges are only to be used once.}$

b. $\text{Minimum distance = 19 m}$

Show Worked Solution

a. $\text{Vertices \(C$ and $E$ are of odd degree.}\)

$\text{The circuit described is Eulerian which requires all vertices to be}$

$\text{of even degree if edges are only to be used once.}$

♦ Mean mark (a) 40%.

b. $\text{Minimum distance}$

$=28+11+23+31+7+7+8+12+22+19+14+14$

$=196 \ \text{m}$

♦♦ Mean mark (b) 27%.

Matrices, GEN2 2025 VCAA 14

An early learning centre runs seven different activities during its 40-day holiday program.

The activities are cooking $(C)$, drama $(D)$, gardening $(G)$, lunch $(L)$, music $(M)$, reading $(R)$ and sport $(S)$.

The timetabled order of the activities for day one of the holiday program is shown below.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ \text{9 am} \ \ \rule[-1ex]{0pt}{0pt} & \text{10 am} \rule[-1ex]{0pt}{0pt} & \text{11 am} \rule[-1ex]{0pt}{0pt} & \text{12 pm} \rule[-1ex]{0pt}{0pt} & \ \ \text{1 pm} \ \ \rule[-1ex]{0pt}{0pt} & \ \ \text{2 pm} \ \ \rule[-1ex]{0pt}{0pt} & \ \ \text{3 pm} \ \ \\
\hline
\rule{0pt}{2.5ex} \textit{C} \rule[-1ex]{0pt}{0pt} & \textit{D} \rule[-1ex]{0pt}{0pt} & \textit{G} \rule[-1ex]{0pt}{0pt} & \textit{L} \rule[-1ex]{0pt}{0pt} & \textit{M} \rule[-1ex]{0pt}{0pt} & \textit{R} \rule[-1ex]{0pt}{0pt} & \textit{S}\\
\hline
\end{array}

The timetabled order of the activities for day one is also shown in matrix $X$ below.

\begin{aligned}
X = & \begin{bmatrix}
C \\
D \\
G \\
L \\
M \\
R\\
S \\
\end{bmatrix}
\end{aligned}

Matrix $P$, shown below, is a permutation matrix used to determine the timetabled order of activities from one day to the next.

\begin{aligned}
P = & \begin{bmatrix}
0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{aligned}

A column matrix containing the timetabled order of activities on one day is multiplied by matrix $P$ to determine the timetabled order of activities for the next day.

State the activities that are always held at the same time on each day of the program. (1 mark)

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Determine the timetabled order of the seven activities on day three of the program. (1 mark)

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$P^4$ is an identity matrix.
Explain what this means for the timetabled order of the activities over the 40-day holiday program. (1 mark)

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a. $\text{Gardening, lunch, music.}$

b. $\text{Drama, cooking, gardening, lunch, music, sport, reading.}$

c. $\text{Order of activities rotate on a 4 day cycle.}$

$\text{Over 40 days, there will be 10 cycles of activities}$

Show Worked Solution

a. $\text{Activities held af the same time:}$

$\Rightarrow \ \text{correspond to 1’s in leading diagonal}$

$\text{Gardening, lunch, music.}$

\begin{aligned}X_2=P \times X=\begin{bmatrix}
R \\
S \\
G \\
L \\
M \\
D \\
C
\end{bmatrix}, \quad X_3=P \times X_2=\begin{bmatrix}
D \\
C \\
G \\
L \\
M \\
S \\
R
\end{bmatrix}
\end{aligned}

$\text{Order on day 3:}$

$\text{Drama, cooking, gardening, lunch, music, sport, reading.}$

♦♦ Mean mark (b) 27%.

c. $P^4 \ \Rightarrow \ \text{identity matrix}$

$\text{Order of activities rotate on a 4 day cycle.}$

$\text{Over 40 days, there will be 10 cycles of activities}$

♦♦♦ Mean mark (c) 11%.

Networks, GEN2 2025 VCAA 15

Frances lives in a housing estate.

On the graph below the vertices represent her favourite locations, and the edges represent the roads between them.

Calculate the sum of the degrees of all the vertices in this graph. (1 mark)

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Euler's formula, $v+f=e+2$, holds for this graph.
Complete the formula by writing the appropriate numbers in the boxes below. (1 mark)

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Frances is at the gym. She would like to visit each of the other locations once and end at her home.
What is the mathematical term used to describe this route? (1 mark)

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Using edges from the original graph, construct a spanning tree below. (1 mark)

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

b. $5+4=7+2$

c. $\text{Hamiltonian path}$

d. $\text{Spanning tree (one of many possibilities):}$

Show Worked Solution

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

b. $v+f=e+2 \ \ \Rightarrow \ \ 5+4=7+2$

c. $\text{Hamiltonian path}$

♦ Mean mark (c) 53%.

d. $\text{Spanning tree (one of many possibilities):}$

Matrices, GEN2 2025 VCAA 12

The early learning centre contains three rooms, Nursery $(N)$, Toddler $(T)$ and Pre-kinder $(P)$ .

From one year to the next, children can move between rooms, stay in the same room, or may leave $(L)$ the centre. The following transition matrix, $M$, shows the expected proportion of children who will move between categories or stay in the same category from one year to the next.

\begin{aligned}
& \quad \quad \quad \quad \quad \textit{this year} \\
& \quad \quad \ \ \ \ N \quad \quad \ \ T \quad \quad P \quad \ L \\
M&=\begin{bmatrix}
0.25 & 0 & 0 & 0 \\
0.625 & 0.25 & 0 & 0 \\
0 & 0.625 & 0.1 & 0 \\
0.125 & 0.125 & 0.9 & 1
\end{bmatrix}\begin{array}{l}
N\\
T \\
P \\
P
\end{array}\quad \textit{next year}
\end{aligned}

The number of children expected to be in each of the four categories, from one year to the next, can be calculated using the matrix recurrence relation
1. $S_{n+1}=M S_n$
where $S_n$ represents the expected number of children in each of the four categories at the start of year $n$.
The state matrix $S_{2024}$, shown below, gives the number of children in each category at the start of 2024.
1. 1. \begin{align*}S_{2024}=\left[\begin{array}{c}4 \\15 \\15 \\27\end{array}\right]\begin{aligned}& N \\& T \\& P \\& L\end{aligned}\end{align*}
Find $S_{2023}$. (1 mark)

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From the start of 2025, new children commenced in the early learning centre at the start of each year.
A new matrix recurrence relation for determining the expected number of children in each of the four categories from one year to the next is
1. 1. \begin{align*}
    S_{n+1}=M S_n+B
    \end{align*}
where
1. 1. \begin{align*}B=\left[\begin{array}{c}12 \\5 \\10 \\0\end{array}\right] \begin{aligned}& N \\& T \\& P \\& L\end{aligned}\end{align*}
gives the number of new children enrolled in each room of the early learning centre at the start of each year.
Given the state matrix
1. \begin{align*}S_{2024}=\left[\begin{array}{c}4 \\15 \\15 \\27\end{array}\right]\begin{aligned}& N \\& T \\& P \\& L\end{aligned}\end{align*}
find the expected total number of children to be enrolled in the early learning centre at the start of 2026. Round your answer to the nearest whole number. (1 mark)

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$S_{2023}=\begin{bmatrix}16 \\ 20 \\ 25 \\ 0\end{bmatrix}$

b. $\text{Children enrolled (2026) =50}$

Show Worked Solution

a. $S_{n+1}=M S_n \ \Rightarrow \ S_n=M^{-1} S_{n+1}$

$S_{2023}=\begin{bmatrix}0.25 & 0 & 0 & 0 \\ 0.625 & 0.25 & 0 & 0 \\ 0 & 0.625 & 0.1 & 0 \\ 0.125 & 0.125 & 0.9 & 1\end{bmatrix}^{-1}\begin{bmatrix}4 \\ 15 \\ 15 \\ 27\end{bmatrix}=\begin{bmatrix}16 \\ 20 \\ 25 \\ 0\end{bmatrix}$

♦ Mean mark (a) 53%.

$S_{2025}=M S_{2024} + B=\begin{bmatrix}13 \\ 11.25 \\ 20.87 \\ 42.87\end{bmatrix}$

$S_{2026}=M S_{2025} + B=\begin{bmatrix}15.25 \\ 15.93 \\ 19.11 \\ 64.69\end{bmatrix}$

$\text{Children enrolled (2026)\(=15.25+15.93+19.11=50$ (nearest whole)}\)

♦♦♦ Mean mark (b) 13%.

Financial Maths, STD2 F4 2025 GEN1 7

Declan is a filmmaker and content creator.

He has taken out a reducing balance loan to fund a new production.

Interest is calculated monthly and Declan makes monthly repayments.

Three rows of the amortisation table for Declan’s loan are shown below.

\begin{array}{|c|c|c|c|c|}
\hline
\hline \rule{0pt}{2.5ex}\quad \textbf{Payment} \quad & \quad\textbf{Payment} \quad & \quad\textbf{Interest} \quad& \textbf{Principal} & \quad\textbf{Balance}\quad\\
\textbf{number} & \textbf{(\$)} \rule[-1ex]{0pt}{0pt}& \textbf{(\$)} & \quad\textbf{reduction (\$)} \quad& \textbf{(\$)}\\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 0.00 & 0.00 & 0.00 & 850\,000.00 \\
\hline \hline \rule{0pt}{2.5ex}1\rule[-1ex]{0pt}{0pt} & 15\,730.88 & 2975.00 & 12\,755.88 & 837\,244.12 \\
\hline \hline \rule{0pt}{2.5ex}2 \rule[-1ex]{0pt}{0pt}& 15\,730.88 & 2930.35 & 12\,800.53 & 824\,443.59 \\
\hline
\end{array}

What amount, in dollars, did Declan borrow? (1 mark)
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Why is the interest associated with payment 2 lower than the interest associated with payment 1? (1 mark)
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The interest rate on Declan’s loan is 4.2% per annum, compounding monthly.
Using the values in the table, complete the table below.
Round all values to the nearest cent. (2 marks)

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a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Interest}=\dfrac{4.2}{100} \times \dfrac{1}{12} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

Show Worked Solution

a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Monthly interest}=\dfrac{4.2}{12}=0.35\%$

$\text{Calculating missing values in the table:}$

$\text{Interest}=\dfrac{0.35}{100} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

♦ Mean mark (c) 40%.

Matrices, GEN2 2025 VCAA 11

An early learning centre contains three rooms, Nursery $(N)$, Toddler $(T)$ and Pre-kinder $(P)$.

The Nursery and Toddler rooms each have capacity for eight children and the Pre-kinder room has capacity for 20 children, as shown in matrix $C$ below.

\begin{align*}
C=\left[\begin{array}{c}
8 \\
8 \\
20
\end{array}\right] \begin{aligned}
& N \\
& T \\
& P
\end{aligned}
\end{align*}

Matrix $E$ shows enrolment numbers for each room for one week, Monday to Friday.

\begin{aligned}
& \quad \ \ \ Mon \quad Tue\quad Wed \ \ \ Thu \ \ \ Fri \\
E&=\begin{bmatrix}
6 & \quad 8 & \quad 8 & \quad 8 & \quad 5 \\
7 & \quad 8 & \quad 7 & \quad 8 & \quad 6 \\
18 & \ \ \ 18 &\ \ \ 17 &\ \ \ 15 &\ \ \ 13
\end{bmatrix}\begin{array}{l}
N \\
T \\
P
\end{array}\\
\end{aligned}

State the order of matrix $E$. (1 mark)
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The following matrix multiplication has been completed to determine a new matrix, $W$.
1. \begin{align*}
  \begin{bmatrix}
  1 & 1 & 1
  \end{bmatrix} \times E=W
  \end{align*}
What information does matrix $W$ provide regarding enrolments at the early learning centre? (1 mark)

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It has been decided that the capacity of the Nursery room will be increased by 25% and the capacity of the Toddler room will be increased by 50%. The capacity of the Pre-kinder room will be reduced by 10%.
The new capacities for the three rooms $(C_{\text {new}})$ can be determined from the matrix product
1. $C_{\text {new }}=F C$
where $F$ is a diagonal matrix.
Write down the matrix $F$. (1 mark)
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a. $\text{Order of matrix} \ \ E=3 \times 5$

b. $\text { Total enrolments for each day of the week.}$

$F=\begin{bmatrix}1.25 & 0 & 0 \\ 0 & 1.5 & 0 \\ 0 & 0 & 0.9\end{bmatrix}$

Show Worked Solution

a. $\text{Order of matrix} \ \ E=3 \times 5$

b. $\text { Total enrolments for each day of the week.}$

$F=\begin{bmatrix}1.25 & 0 & 0 \\ 0 & 1.5 & 0 \\ 0 & 0 & 0.9\end{bmatrix}$

♦ Mean mark (c) 44%.

Financial Maths, GEN2 2025 VCAA 9

Declan takes out a new loan of $50 000 to promote his new film.

Interest on this loan compounds weekly and Declan makes weekly repayments of $75.

With these weekly repayments of $75, suppose the balance of Declan’s loan does not change over time.
Determine the weekly interest rate. (1 mark)

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With these weekly repayments of $75, suppose the balance of Declan's loan now reduces over time.
The balance of the loan, in dollars, after $n$ weeks, $L_n$, can be determined using a recurrence relation of the form
1. $L_0=50\,000, \quad L_{n+1}=R L_n-75$
Assume there are exactly 52 weeks in a year
After one year, Declan owes $49 565.34
Determine
1. the per annum interest rate, compounding weekly, as a percentage, rounded to two decimal places. (1 mark)
  
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2. the value of $R$ rounded to four decimal places. (1 mark)
  
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a. $\text{Weekly interest rate}=0.15 \%$

b.i. $\text{Interest rate}=6.96 \% \ \text{(2 d.p)}$

b.ii. $R=1.0013$

Show Worked Solution

a. $\text{Weekly interest rate}=\dfrac{75}{50\,000} \times 100=0.15 \%$

b.i. $L_0=50\,000 \quad L_{n+1}=R L_n-75$

$\text{When} \ \ N=52,\ \text{balance}=\$ 49\,565.34$

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

$N$	$=52$
$I(\%)$	$=\boldsymbol{6.96000}$
$PV$	$=50\,000$
$PMT$	$=-75$
$FV$	$=-49\,565.34$
$PY$	$=CY=52$

$\text{Interest rate}=6.96 \% \ \text{(2 d.p)}$

b.ii. $R=1+\dfrac{6.96}{100 \times 52}=1.0013$

♦ Mean mark (a) 40%.
♦ Mean mark (b.i) 40%
♦♦ Mean mark (b.ii) 30%

Financial Maths, GEN2 2025 VCAA 8

Declan depreciates the value of his lighting equipment using flat rate depreciation.

The graph below shows the value, in dollars, of the lighting equipment, $V_n$, after $n$ years.

The value of the lighting equipment could be modelled by either a recurrence relation or a rule.
1. Write a recurrence relation in terms of $V_0, V_{n+1}$ and $V_n$. (1 mark)
  
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2. Write a rule for $V_n$ in terms of $n$. (1 mark)
  
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What is the annual flat rate depreciation percentage applied to the lighting equipment? (1 mark)
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a.i. $V_0=40\,000 \quad V_{n+1}=V_n-8000$

a.ii. $V_n=40\,000-8000 \times n$

b. $\text{Depreciation}=20\%$

Show Worked Solution

a.i. $\text{Recurrence relation:}$

$V_0=40\,000 \quad V_{n+1}=V_n-8000$

a.ii. $V_n=40\,000-8000 \times n$

b. $\text{Depreciation %}=\dfrac{8000}{40\,000} \times 100=20\%$

♦ Mean mark (a.ii) 50%.

Financial Maths, GEN2 2025 VCAA 7

Declan is a filmmaker and content creator.

He has taken out a reducing balance loan to fund a new production.

Interest is calculated monthly and Declan makes monthly repayments.

Three rows of the amortisation table for Declan’s loan are shown below.

What amount, in dollars, did Declan borrow? (1 mark)
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Why is the interest associated with payment 2 lower than the interest associated with payment 1? (1 mark)
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The interest rate on Declan’s loan is 4.2% per annum, compounding monthly.
Using the values in the table, complete the table below.
Round all values to the nearest cent. (1 mark)

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The last payment required to fully repay the loan is $15 730.71, correct to the nearest cent.
How many payments of $15 730.88 did Declan make before this final payment? (1 mark)

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a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Interest}=\dfrac{4.2}{100} \times \dfrac{1}{12} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

d. $\text{59 payments made before the final payment.}$

Show Worked Solution

a. $\$ 850\,000$

b. $\text{Interest is lower (payment 2 vs payment 1) because it is based on}$

$\text{a reduced balance.}$

c. $\text{Interest}=\dfrac{4.2}{100} \times \dfrac{1}{12} \times 824\, 443.59=\$ 2885.55$

$\text{Principal reduction}=15\,730.88-2885.55=\$ 12\,845.33$

$\text{Balance}=824\,443.59-12\,845.33=\$811\,598.26$

♦ Mean mark (c) 40%.

d. $\text{Solve for \(N$ (by CAS):}\)

$N$	$=\boldsymbol{59.99 \ldots}$
$I(\%)$	$=4.2$
$PV$	$=850\,000$
$PMT$	$=-15\,730.88$
$FV$	$=0$
$PY$	$=CY=12$

$\therefore \ \text{59 payments made before the final payment.}$

♦♦♦ Mean mark (d) 21%.

Data Analysis, GEN2 2025 VCAA 5

The table below shows sale price and the number of days on the market before sale, days, for a sample of 10 apartments sold in a particular suburb.

\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex}\ \ \textbf{Sale price(\$)} \ \ \rule[-1ex]{0pt}{0pt}& \quad \quad \textbf{Days}\quad \quad \\
\hline \rule{0pt}{2.5ex}950\,000 \rule[-1ex]{0pt}{0pt}& 15 \\
\hline \rule{0pt}{2.5ex}925\,000 \rule[-1ex]{0pt}{0pt}\rule[-1ex]{0pt}{0pt}& 18 \\
\hline \rule{0pt}{2.5ex}900\,000 & 23 \\
\hline \rule{0pt}{2.5ex} 900\,000 \rule[-1ex]{0pt}{0pt}& 24 \\
\hline \rule{0pt}{2.5ex}905\,000 \rule[-1ex]{0pt}{0pt}& 26 \\
\hline \rule{0pt}{2.5ex}750\,000 \rule[-1ex]{0pt}{0pt}& 28 \\
\hline \rule{0pt}{2.5ex}680\,000 \rule[-1ex]{0pt}{0pt}& 31 \\
\hline \rule{0pt}{2.5ex}800\,000 \rule[-1ex]{0pt}{0pt}& 35 \\
\hline \rule{0pt}{2.5ex}590\,000 \rule[-1ex]{0pt}{0pt}& 46 \\
\hline \rule{0pt}{2.5ex}600\,000 \rule[-1ex]{0pt}{0pt}& 65 \\
\hline
\end{array}

Use the data in the table above to find the equation of the least squares line.
Write your answers in the boxes below, rounding both values to four significant figures. (2 marks)

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sale price$=\begin{array}{|c|}\hline \rule{0pt}{5 ex} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \hline \end{array} \times \begin{array}{|c|}\hline \rule{0pt}{5 ex}\quad \quad \quad \quad \quad \quad\\ \hline \end{array} \times$days
For this data, Pearson's correlation coefficient is $r=-0.866$, rounded to three decimal places.
Explain the meaning of the coefficient of determination, as a whole percentage, in the context given in this question. (1 mark)

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a. $\textit{sale price}=1\,050\,000-8050 \times \textit{days} \ \text{(4 sig. fig.)}$

b. $\text{75% of the variation in sale price can be explained by the variation in days.}$

Show Worked Solution

a. $\text{By calculator:}$

$\textit{sale price}=1\,050\,000-8050 \times \textit{days} \ \text{(4 sig. fig.)}$

♦ Mean mark (a) 50%.

b. $(-0.866)^2=0.7499 \ldots \approx 75 \%$

$\text{75% of the variation in sale price can be explained by the variation in days.}$

♦ Mean mark (b) 44%.

Data Analysis, GEN2 2025 VCAA 4

The scatterplot below shows the sale price of a home, in dollars, against the distance of the home from the city centre of Melbourne, in kilometres, distance from city centre.

The sample consists of three‑bedroom homes sold between 2016 and 2018

The equation of the least squares line for the data in the scatterplot is

sale price$=1\,765\,353-35\,054 \times$distance from city centre

The coefficient of determination is 0.0806

Identify the explanatory variable in the least squares equation. (1 mark)

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Calculate the value of the correlation coefficient $r$. Round your answer to three decimal places. (1 mark)

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Use the equation of the least squares line to predict the sale price for a three-bedroom home, located in the city centre of Melbourne, sold between 2016 and 2018. (1 mark)

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Jocelyn wants to sell her three-bedroom home located two kilometres from the city centre of Melbourne.
Would the predicted sale price be an example of interpolation or extrapolation?
Briefly explain your answer. (1 mark)

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Describe the linear association between sale price and distance from city centre in terms of its strength and direction. Answer in the table below. (2 marks)

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\begin{array}{|l|l|}
\hline \rule{0pt}{2.5ex}\text {strength} \quad \quad \rule[-1ex]{0pt}{0pt}& \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\
\hline \rule{0pt}{2.5ex}\text {direction} \rule[-1ex]{0pt}{0pt}& \\
\hline
\end{array}

A residual plot associated with the least squares line is shown below.
It is missing one point.

The residual associated with the home that is furthest from the city centre of Melbourne is missing from the residual plot. The home is 15.5 km from the city centre and sold for $1 250 000.
1. Show that the value of the missing residual is 27 984. (1 mark)
  
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2. Plot the residual from part i by placing an $\text{X}$ on the residual plot above. (1 mark)
  
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Show Answers Only

a. $\text{Distance from city centre}$

b. $r=-0.284$

c. $\text{sale price}=1\,765\,353$

d. $\text{Extrapolation as 2 km lies outside the explanatory variable data range.}$

e. $\text{Strength: weak. Direction: negative}$

f.i. $\text{Sale price (est)}=1\,765\,353-35\,054 \times 15.5=1\,222\,016$

$\text{Actual sale price}=1\,250\,000$

$\text{Residual}=1\,250\,000-1\,222\,016=27\,984$

f.ii.

Show Worked Solution

a. $\text{Distance from city centre}$

b. $\text{Since slope is negative}$

$r=-\sqrt{0.0806}=-0.284$

♦♦♦ Mean mark (b) 21%.

c. $\text{Find sale price when distance fran city centre}=0:$

$\text{sale price}=1\,765\,353$

d. $\text{Extrapolation as 2 km lies outside the explanatory variable data range.}$

$\text{(Note: interpolation/extrapolation should be referenced to the}$

$\text{explanatory variable range).}$

♦♦ Mean mark (d) 27%.

e. $\text{Strength: weak}$

$\text{Direction: negative}$

f.i. $\text{Sale price (est)}=1\,765\,353-35\,054 \times 15.5=1\,222\,016$

$\text{Actual sale price}=1\,250\,000$

$\text{Residual}=1\,250\,000-1\,222\,016=27\,984$

♦ Mean mark (f.i) 41%.
♦ Mean mark (f.ii) 45%

f.ii.

Data Analysis, GEN2 2025 VCAA 1

Table 1, below, shows the prices in dollars, price, for a sample of 20 homes sold in an inner Melbourne suburb during 2017.

The type of home sold is either an apartment or a house.

Table 1

\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex}\quad \ \textit{Price(\$)}\quad \ \rule[-1ex]{0pt}{0pt}& \textit{Type} \\
\hline \rule{0pt}{2.5ex}350\,000 \rule[-1ex]{0pt}{0pt}& \ \ \text{apartment}\ \ \\
\hline \rule{0pt}{2.5ex}490\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}500\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}620\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}720\,000 \rule[-1ex]{0pt}{0pt}\rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}830\,000 & \text{apartment}\\
\hline \rule{0pt}{2.5ex}875\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}995\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}1\,100\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}1\,520\,000 \rule[-1ex]{0pt}{0pt}& \text{apartment}\\
\hline \rule{0pt}{2.5ex}800\,000 \rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}840\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}920\,000 \rule[-1ex]{0pt}{0pt}& \text{house} \\
\hline \rule{0pt}{2.5ex}920\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,010\,000\rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}1\,263\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,398\,000 \rule[-1ex]{0pt}{0pt}& \text{house}\\
\hline \rule{0pt}{2.5ex}1\,460\,000\rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline \rule{0pt}{2.5ex}1\,540\,000 \rule[-1ex]{0pt}{0pt}& \text{house} \\
\hline \rule{0pt}{2.5ex}1\,540\,000 \rule[-1ex]{0pt}{0pt} & \text{house}\\
\hline
\end{array}

Find the median, in dollars, of the variable price. (1 mark)

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State whether the variable type is numerical, nominal or ordinal. (1 mark)

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1. Complete the table below by finding the standard deviation, to the nearest whole number, for the sale price of apartments in the sample. (1 mark)
  
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2. Table 2
  \begin{array}{|c|c|}
  \hline \rule{0pt}{2.5ex} \quad \ \ \textbf{Type of home} \quad \ \ & \quad \textbf{Standard deviation of} \quad \\
  & \rule[-1ex]{0pt}{0pt}\textbf{sale price (\$)}\\
  \hline \rule{0pt}{2.5ex}\text{house} \rule[-1ex]{0pt}{0pt}& 300\,911 \\
  \hline \rule{0pt}{2.5ex}\text{apartment} \rule[-1ex]{0pt}{0pt}& \\
  \hline
  \end{array}
3. Using the information in Table 2, comment on the relative spread in the distribution of the sale prices of houses compared with apartments in this sample. (1 mark)
  
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Table 3, below, shows the percentage of houses and apartments with prices in the given ranges. Some information is missing.
Use the data from Table 1 to complete Table 3.
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Table 3

Show Answers Only

a. $\text{Median}=920\,000$

b. $\text{Variable is nominal}$

c.i. $\text{Std deviation = \$346 466}$

c.ii. $\text {Apartment sale prices have a higher spread (standard deviation) than the}$

$\text{spread of house sale prices.}$

\begin{array}{|c|c|}
\hline \hline \rule{0pt}{2.5ex}\ \ \ \text {House}(\%) \ \ \ \rule[-1ex]{0pt}{0pt}& \text {Apartment}(\%) \\
\hline \hline \rule{0pt}{2.5ex}0 \rule[-1ex]{0pt}{0pt}& 30 \\
\hline \hline \rule{0pt}{2.5ex}40 \rule[-1ex]{0pt}{0pt}& 50 \\
\hline \hline \rule{0pt}{2.5ex}60 \rule[-1ex]{0pt}{0pt}& 20 \\
\hline \hline \rule{0pt}{2.5ex}100 \rule[-1ex]{0pt}{0pt}& 100 \\
\hline
\end{array}

Show Worked Solution

a. $\text{Median}=\dfrac{10^{\text{th}}+11^{\text{th}}}{2}=920\,000$

b. $\text{Type is qualitative and cannot be ordered}$

$\Rightarrow \ \text{Variable is nominal}$

c.i. $\text{By calculator,}$

$\text{Std deviation = \$346 466}$

c.ii. $\text {Apartment sale prices have a higher spread (standard deviation) than the}$

$\text{spread of house sale prices.}$

♦ Mean mark (c.ii) 47%.

Networks, STD2 N3 EQ-Bank 38

The precedence table below shows the 12 activities required to complete a project. The duration in days and immediate predecessors are shown.

\begin{array}{|c|c|c|}
\hline \rule{0pt}{2.5ex}\ \ \textbf{Activity} \ \ & \ \ \textbf{Duration} \ \ & \textbf{Immediate predecessors} \\
\hline \rule{0pt}{2.5ex}A \rule[-1ex]{0pt}{0pt}& 4 & - \\
\hline \rule{0pt}{2.5ex}B \rule[-1ex]{0pt}{0pt}& 6 & A \\
\hline \rule{0pt}{2.5ex}C \rule[-1ex]{0pt}{0pt}& 8 & A \\
\hline \rule{0pt}{2.5ex}D \rule[-1ex]{0pt}{0pt}& 3 & A \\
\hline \rule{0pt}{2.5ex}E \rule[-1ex]{0pt}{0pt}& 9 & B\\
\hline \rule{0pt}{2.5ex}F \rule[-1ex]{0pt}{0pt}& 6 & C \\
\hline \rule{0pt}{2.5ex}G \rule[-1ex]{0pt}{0pt}& 7 & B, D, F \\
\hline \rule{0pt}{2.5ex}H \rule[-1ex]{0pt}{0pt}& 12 & C \\
\hline \rule{0pt}{2.5ex}I \rule[-1ex]{0pt}{0pt}& 6 & G, H \\
\hline \rule{0pt}{2.5ex}J \rule[-1ex]{0pt}{0pt}& 4 & E, I \\
\hline \rule{0pt}{2.5ex}K \rule[-1ex]{0pt}{0pt}& 3 & G, H \\
\hline \rule{0pt}{2.5ex}L \rule[-1ex]{0pt}{0pt}& 9 & J \\
\hline
\end{array}

The project is to be completed in minimum time.

Sketch the network, identifying each activity and its duration, including any dummy activities. (3 marks)
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Determine the critical path of the network. (1 mark)
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Show Answers Only

a. $\text{Sketch the network:}$

b. $\text {Critical Path:}\ ACFGIJL$

Show Worked Solution

a. $\text{Sketch the network:}$

b. $\text{Critical path only requires the forward scanning in the above network.}$

$\text {Critical Path:}\ ACFGIJL$

Networks, GEN1 2025 VCAA 40 MC

The precedence table below shows the 12 activities required to complete a project. The duration in days and immediate predecessors are shown.

The project is to be completed in minimum time.

The float time, in days, of Activity $B$ is

Show Answers Only

$C$

Show Worked Solution

$\text{The precedence table produces the network:}$

$\text {Critical Path:}\ ACFGIJL$

$\text{Float time of \(B=$ LST of $E-$EST of $B-$duration $B=18-4-6=8$}\)

$\Rightarrow C$

♦♦ Mean mark 39%.

Networks, GEN1 2025 VCAA 38 MC

The diagram below shows the network of roads that Sofia can use to travel between home and school.

The numbers on the roads show the length, in metres, of each section of road.

Using this network of roads, the shortest distance that Sofia can take to travel from home to school, in metres, is

3700
3800
3900
5100

Show Answers Only

$A$

Show Worked Solution

$\text{Shortest path}=900+500+400+800+1100=3700$

$\Rightarrow A$

♦ Mean mark 47%.

\(\operatorname{Pr}(X \geqslant 3)\)	\(=\operatorname{Pr}(X=3)+\operatorname{Pr}(X=4)\)
	\(=4 p^3(1-p)+p^4\)
	\(=4 p^3-4 p^4+p^4\)
	\(=4 p^3-3 p^4\)
	\(=g(p)\)

\(\operatorname{Pr}(X \geqslant 3)\)	\(=\operatorname{Pr}(X=3)+\operatorname{Pr}(X=4)\)
	\(=4 p^3(1-p)+p^4\)
	\(=4 p^3-4 p^4+p^4\)
	\(=4 p^3-3 p^4\)
	\(=g(p)\)

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

\(\displaystyle k \int_0^{\tfrac{\pi}{4}} \sin x+k \int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} \cos x\)	\(=1\)
\(k[-\cos x]_0^{\tfrac{\pi}{4}}+k[\sin x]_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\)	\(=1\)
\(k\left(-\dfrac{1}{\sqrt{2}}+1\right)+k\left(1-\dfrac{1}{\sqrt{2}}\right)\)	\(=1\)

\(k\)	\(=\dfrac{\sqrt{2}}{2 \sqrt{2}-2} \times \dfrac{2 \sqrt{2}+2}{2 \sqrt{2}+2}\)
	\(=\dfrac{4+2 \sqrt{2}}{8-4}\)
	\(=1+\dfrac{1}{2} \sqrt{2}\)

\(y=g(f(x))\)	\(=-x(-x+1)(-x-1)(-x-2) \)
	\(=x(1-x)(x+1)(-x-2) \)
	\(=-x(1-x)(x+1)(x+2) \)

\(\displaystyle \int \frac{1}{y^3}\ d y\)	\(=\displaystyle \int x^2\ d x\)
\(-\dfrac{1}{2 y^2}\)	\(=\dfrac{1}{3} x^3+c\)

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

\(7-6 x^3\)	\(>0\)
\(6 x^3\)	\(<7\)
\(x^3\)	\(<\dfrac{7}{6}\)
\(x\)	\(< \sqrt[3]{\dfrac{7}{6}}\)